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ESSENTIALS OF ALGEBRA

SECONDARY SCHOOLS

BY

WEBSTER WELLS, S.B.

PROFESSOR OF MATHEMATICS IN THE MASSACHUSETTS
INSTITUTE OF TECHNOLOGY

o>IHc

D. C. HEATH & CO., PUBLISHERS
BOSTON NEW YORK CHICAGO

Copyright, 1897,

Bt webstee wells.

PREFACE.

The cordial reception which the author's other Algebras
have received at the hands of the educational public, their
extensive use in schools of the highest rank in all parts of
the country, the appreciative recommendations which have
come to him from instructors of reputation, lead him to
believe that this latest attempt to adequately meet the
demands of the best secondary schools will be cordially
welcomed.

Our teachers are progressive, and the author who fails to
keep abreast of the times, and in sympathy with the best
educational thought and methods, will appeal in vain for
the patronage and sympathies of his fellow-teachers.

Fully conscious of the above truth, the author earnestly
recommends " The Essentials of Algebra " to the attention
of the educational public.

It affords a thorough and complete treatment of elemen-
tary Algebra, and attention is especially invited to the fol-
lowing features : —

The introduction of easy problems at the very outset ; § 5.

The Addition and Multiplication of Positive and Nega-
tive Numbers ; §§ 14 to 19.

The Addition of Similar Terms ; § 31.

The discussion of Simple Equations, not involving Frac-
tions, directly after Division ; Chap. VII.

The suggestions in regard to the solution of problems ;
§§ 76, 77.

The discussion of the theoretical principles involved in
the handling of fractions ; §§ 129, 136, 143, 145.

iv PREFACE.

The examples on page 176.

The discussion of square roots and cube roots of arith-
metical numbers; §§ 197, 198, 203, 204.

The examples at the end of § 229.

The solution of equations by factoring ; §§ 266, 267.

The factoring of a quadratic expression when the co-
efficient of x^ is a perfect square ; § 286.

Great care has been taken to state the various definitions
and rules with accuracy, and every principle has been dem-
onstrated with strict regard to the logical principles in-
volved. As a rule, no definition has been introduced until
its use became necessary.

The examples and problems have been selected with great
care, are ample in number, and thoroughly graded. They
are especially numerous in the important chapters on Fac-
toring, Fractions, and Eadicals.

The latest English practice has been followed in writing
Arithmetic, Geometric, and Harmonic, for Arithmetical,
Geometrical, and Harmonical, in the progressions.

The author wishes to acknowledge, with hearty thanks,
the many suggestions and the assistance that he has received
from principals and teachers of secondary schools in all
parts of the country, in improving and perfecting the
work.

WEBSTER WELLS.

Massachusetts Institute of Technology,
March, 1897.

OOITTEJSrTS.

PAGE

I. Definitions and Notation 1

Solution of Problems by Algebraic Methods .... 2

Algebraic Expressions 6

II. Positive and Negative Numbers 9

Addition of Positive and Negative Numbers .... 11

Multiplication of Positive and Negative Numbers . . 12

III. Addition and Subtraction of Algebraic Expressions . 15

Addition of Monomials 16

Addition of Polynomials 20

Subtraction 21

Subtraction of Monomials 22

Subtraction of Polynomials ' 23

IV. Parentheses 26

Removal of Parentheses 26

Introduction of Parentheses 28

V. Multiplication 29

Multiplication of Monomials 30

Multiplication of Polynomials by Monomials .... 32

Multiplication of Polynomials by Polynomials ... 32

VI. Division 37

Division of Monomials 38

Division of Polynomials by Monomials 39

Division of Polynomials by Polynomials 40

VII. Simple Equations 48

Properties of Equations 49

Solution of Simple Equations 50

Problems ^2

VIII. Important Rules in Multiplication and Division . . 59

IX. Factoring ^^

V

Vi CONTENTS.

PAGE

X. Highest Common Factor 81

XI. Lowest Common Multiple 91

XII. Fractions 96

Keduction of Fractions 97

Addition and Subtraction of Fractions 105

Multiplication of Fractions Ill

Division of Fractions 113

Complex Fractions 115

XIII. Simple Equations (Continued) 120

Solution of Equations containing Fractions . . . 120

Solution of Literal Equations 124

Solution of Equations involving Decimals .... 126

Problems ^^ 127

Problems involving Literal Equations 135

XIV. Simultaneous Equations

Containing Two Unknown Quantities 138

XV. Simultaneous Equations

Containing more than Two Unknown Quantities. . 150

XVI. Problems

Involving Simultaneous Equations 154

XVII. Inequalities 165

XVIII. Involution 170

Involution of Monomials 170

Square of a Polynomial 171

Cube of a Binomial 172

XIX. Evolution 174

Evolution of Monomials 174

Square Root of a Polynomial 176

Square Root of an Arithmetical Number . , . . 179

Cube Root of a Polynomial 183

Cube Root of an Arithmetical Number 186

XX. Theory of Exponents 191

XXI. Radicals 201

Reduction of a Radical to its Simplest Form . . . 201

Addition and Subtraction of Radicals 205

To Reduce Radicals of Different Degrees to Equiva-
lent Radicals of the Same Degree 206

CONTENTS.

VU

XXI. Radicals (^Continued).. page

Multiplication of Radicals 207

Division of Radicals 210

Involution of Radicals 212

Evolution of Radicals 212

To Reduce a Fi'action liaving an Irrational Denom-
inator to an Equivalent Fraction whose Denom-
inator is Rational 213

Properties of Quadratic Surds .215

Imaginary Numbers 218

Solution of Equations containing Radicals . . . 222

XXII. Quadratic Equations 224

Pure Quadratic Equations 224

Affected Quadratic Equations . 226

Problems 238

XXIII. Equations Solved like Quadratics 243

Equations in the Quadratic Form 243

XXIV. Simultaneous Equations

Involving Quadratics 248

Problems 258

XXV. Theory of Quadratic Equations 261

Factoring 263

Discussion of the General Equation 268

XXVI. Zero and Infinity 270

Variables and Limits 270

The Problem of the Couriers 272

XXVII. Indeterminate Equations 275

XXVIII. Ratio and Proportion 278

Properties of Proportions 279

XXIX. Variation 287

XXX. Progressions 291

Arithmetic Progression 291

Geometric Progression 299

Harmonic Progression 307

XXXI. The Binomial Theorem 310

Positive Integral Exponent 310

viii CONTENTS.

PAOB

XXXII. Undetermined Coefficients 317

Convergency and Divergency of Series .... 317

The Theorem of Undetermined Coefficients ... 320

Expansion of Fractions into Series 321

Expansion of Radicals into Series 323

Partial Fractions 324

Reversion of Series 330

XXXIII. The Binomial Theorem 332

Fractional and Negative Exponents 332

XXXIV. Logarithms 339

Properties of Logarithms 341

Use of the Table 346

Applications 351

Arithmetical Complement 353

Exponential Equations 357

Answers to the Examples.

ALGEBRA.

I. DEFINITIONS AND NOTATION.

1. In Algebra, the operations of Aritlimetic are abridged
and generalized by means of Symbols.

2. Symbols which represent Numbers.

Tlie symbols generally employed to represent numbers
are the Jigures of Arithmetic and the letters of the Alphabet.

Known Numbers are usually represented by the first
letters of the alphabet, as a, b, c.

Unknown Numbers, or those whose values are to be
determined, are usually represented by the last letters
of the alphabet, as x, y, z.

3. Symbols which represent Operations.

The following symbols have the same meaning in Alge-
bra as in Arithmetic :

-}-, read "pZ;<s."

— , read " rnimis."

X, read "times," "into," or "multiplied by."

-7-, read "divided by."

The sign of multiplication is usually omitted in Algebra,
except between arithmetical figures.

Thus, 2 X a? is written 2 x.

Division is usually indicated by a horizontal line.

Thus, a -^ 6 is written -•

0

1

2 ALGEBRA.

4. The Sign of Equality, =, is read ^'equals,'" or "is
equal to.'"

An Equation is a statement that two numbers are equal.

SOLUTION OF PROBLEMS BY ALGEBRAIC METHODS.

5, The following examples will illustrate the use of
Algebraic symbols in the solution of problems.

The utility of the process consists in the fact that the
unknown numbers are represented by symbols, and that
the various operations are stated in Algebraic language.

1. The sum of two numbers is 30, and the greater exceeds
the less by 4 ; what are the numbers ?

We will represent the less number by x.
Then the greater will be represented by x + 4.
By the conditions of the problem, the sum of the greater number
and the less is 30 ; this is stated in Algebraic language as follows :

X + 4 + X = 30. (1)

But the sum of x and x is twice x, or 2x; whence, equation (1)

may be written

2 X + 4 = 30.

Now if 2 X plus 4 equals 30, 2 x must equal 30 — 4, or 26.

Whence, 2 x = 26.

But if twice x is 26, x must be one-half of 26, or 13.

Hence, the less number is 13, and the greater is 13 + 4, or 17.

The written work will stand as follows :

Let X = the less number.

Then, x + 4 = the greater number.

By the conditions, x + 4 + x = 30.

Or, . 2x + 4 = 30.

Whence, 2x = 26.

Dividing by 2, x = 13, the less number.

Whence, x + 4 = 17, the greater number.

DEFINITIONS AND NOTATION. 3

2. The sum of the ages of A and B is 109 years, and A
is 13 years younger than B ; find their ages.

Let X represent the number of years in B's age.
Then, x — 13 will represent the number of years in A's a^e.
By the conditions of the problem, the sum of the ages of A and B
is 109 years.

Whence, a; + a; - 13 = 109.

Or, 2 a; - 13 = 109.

Now if 2 X minus 13 equals 109, 2 x must equal 109 + 13, or 122.

Wlience, ' 2x= 122.

Dividing by 2, a; = 61, the number of years in B's age.

And, a; — 13 = 48, the number of years in A's age.

The written work will stand as follows :

Let X = the number of years in B's age.

Then, x — 13 = the number of years in A's age.

By the conditions, a; + x — 13 = 109.

Or, 2x-13=:109.

Whence, 2 x = 122.

Dividing by 2, x = 61, the number of years in B's age.

Therefore, x — 13 = 48, the number of years in A's age.

3. A, B, and C together have % 66. A has one-half as
much as B, and C has 3 times as much as A. How much
has each?

Let X = the number of dollars A has.

Then, 2 x = the number of dollars B has,

and 3 X = the number of dollars C has.

By the conditions,

x + 2x + 3x = 66.

But the sum of x, twice x, and 3 times x is 6 times x, or 6 x.

Whence, 6x = 66.

Dividing by 6, x = 11, the number of dollars A has.

Whence, 2 x = 22, the number of dollars B has,

and 3 X = 33, the number of dollars C has.

4 ALGEBRA-

PROBLEMS.

4. The greater of two numbers is 4 times the less, and
their sum is 70. What are the numbers ?

5. The sum of the ages of A and B is 116 years, and A
is 18 years younger than B. What are their ages ?

6. Divide 123 into two parts, such that the greater
exceeds the less by 67.

7. The sum of the ages of A and B is 102 years, and A
is 26 "years older than B. What are their ages ?

8. Divide $ 93 between A and B, so that A may receive

$ 23 less than B.

9. Divide $ 56 between A and B, so that A may receive
6 times as much as B.

10. Divide 85 into two parts, one of which shall be 19
less than the other.

11. Divide $ 72 between A and B, so that A may receive
one-third as much as B.

12. A certain hall contains 425 persons ; there are 3 times
as many men as women, and 4 times as many women as
children. How many are there of each?

13. A man had $4.95. After spending a certain sum,
he found that he had left 4 times as much as he had spent.
How much did he spend ?

14. A, B, and C together have $96. B has twice as
much money as C, and A has as much as B and C together.
How much has each ?

15. The sum of three numbers is 168. The second is 23
less than the first, and the third is 3 times the second.
What are the numbers ?

16. A, B, and C together have $ 230. A has $ 21 more
than B. and $ 17 less than C. How much has each ?

DEFINITIONS AND NOTATION. 5

17. A watch and chain are together worth $ 56, and the
chain is worth one-sixth as much as the watch. What is
the value of each ?

18. Divide 169 into three parts, the first of which is one-
half of the second, and the second one-fifth of the third.

19. Divide $ 144 into three parts such that the secoM
is one-third of the first, and one-fourth of the third.

20. A man bought a cow, a sheep, and a hog for $84.
The price of the hog was one-fifth the price of the cow, and
$ 7 less than the price of the sheep. What was the price
of each ?

21. The sum of three numbers is 127. The first is one-
half of the third, and 17 greater than the second. What
are the numbers ?

22. At a certain election, two candidates, A and B, to-
gether received 508 votes; and A had a majority of 136.
How many did each receive ?

23. The sum of the ages of A, B, and C is 101 years.
A is 17 years younger than B, and 15 years older than G.
What are their ages ?

24. Divide $ 174 among A, B, and C, so that A may
receive 4 times as much as B, and $ 42 more than C.

25. My horse, carriage, and harness are together worth
$ 456. The carriage is worth 8 times as much as the har-
ness, and $ 48 less than the horse. Find the value of each.

26. Divide $ 155 into three parts such that the first shall
be 5 times the second, and one-fifth of the third.

27. At a certain election, three candidates, A, B, and C,
together received 512 votes. A received 28 less than B,
and 64 less than C. How many did each receive ?

/ 28. Divide $ 69 among A, B, C, and D, so that A may
receive $ 5 more than B, C as much as A and B together,
and D as much as A and C together.

6 ALGEBRA.

29. The sum of four numbers is 160. The first is 3
times the second, the second 3 times the third, and the
third 3 times the fourth. What are the numbers ?

DEFINITIONS.

6. If a number be multiplied by itself any number of
times, the result is called a power of that number.

An Exponent is a number written at the right of, and
above another number, to indicate what power of the latter
is to be taken.

Thus,
a?, read " a square,^^ or " a second power" denotes a xa;
a^, read "a cube," or "a third jyower," denotes a x a x a;
a*, read " a foia'th," or " a fourth power" denotes a x a X (* x a,
and so on.

If no exponent is expressed, the first power is understood.

Thus, a is the same as a)-.

7. Symbols of Aggregation.

The parentheses ( ), the brackets [ ], the braces { |, and

the vinculum , indicate that the numbers enclosed by

them are to be taken collectively ; thus,

(a + b) X c, [a 4- 6] X c, ^a + blx c, and a + b x c

all indicate that the result obtained by adding 6 to a is
to be multiplied by c.

ALGEBRAIC EXPRESSIONS.

8. An Algebraic Expression, or simply an Expression, is

a number expressed in algebraic symbols ; as,

2, a, OT 2 3:^ -Sab + 5.

The Numerical Value of an expression is the result
obtained by substituting particular numerical values for
the letters involved in it, and performing the operations
indicated.

DEFINITIONS AND NOTATION.

1. Find the nnmerical value of tlie expression

b

when a = 4, 6 = 3, c = 5, and d = 2.

We have, 4a +^- # = 4 x 4 +5-4^-2«

0 o

= 16 + 10 - 8 = 18, Ans.

EXAMPLES.

Find the numerical value of each of the following when
a = 3, b — 5, c = 2, d = 4, m = 4, and n = 3:

8. a'^'d".

a ct , b C

y. - H

b c a

10. 8 a' -de.

u. 1+Ul+l.

abed

12 ii^— ^^^^
' 2c2 4:d''

'' 3 6"' ■ b''^ c" d^'

If the expression involves parentheses, the operations
indicated ivithin the parentheses must be performed first.

14. Find the numerical value, when a = 9, 6 = 7, and
c = 4, of

(„_5)(6 + o)-|±|.

We have, a - b = 2, 6 + c = 11, a + 5 = 16, and 6 - c = 3.
Then the numerical value of the expression is

2xll-i? = 22-^ = ^. Ans.

2.

ad' - 6c3.

3.

3 abed.

4.

4 a^fZ — 5 6c -

-6cd.

5.

a6 cd
c a

6.

ft

a b c

5 a"

8 ALGEBRA.

Find the numerical value of each of the following expres-
sions, when a = 5, 6 = 3, c = 4, and d = 2 :

15. (^-|Y- 17. 3aXc-d)-2b\c + d).

16. ia' + b' - (f)^. 18. 5(a + by-9{c-(T)K

19. (a-b)(b + c)-(c-d)(d + a).

20. (a-b + c-d)(a-\-b-c-cb.

2 J 3a — 55 -f- 7c gg a + b b j- c c + d

'4a — 36-f6c "64-c c + d d + a

Find the numerical value of each of the following expres-
sions, when a = -f, & = ^, c = |, and a; = 3 :

23 tt + & a — 5^ 24 16 a — 18 6 + 15 c

' a — b a + b ' U a — 32 b — 27 c

25. ^-3 - (4 a - 5 c) a^ + (2 a + 6)a; - 12 abc

\,& aj \a b cj c2

9. Axioms.

An Axiom is a truth which is assumed as self-evident.

Algebraic operations are based upon the following axioms :

1. If the same oj^eration be performed upon equal numbers,
the resulting numbers will be equal.

2. If the same number be both added to, and subtracted
from another, the value of the latter loill not be changed.

3. If a number be both midtiplied and divided by another,
the value of the former will not he changed.

4. Numbers which are equal to the same number, are equal
to ea^h other.

POSITIVE AA'D NEGAllVE i^UAlliEiib.

II. POSITIVE AND NEGATIVE NUMBERS.

10. Many concrete magnitudes are capable of existing in

two opposite states.

Thus, in financial transactions, we may have gains, or
losses; in the thermometer, we may have temperatures above
zero, or beloio zero ; a place on the surface of the earth may
be in north latitude, or south latitude ; etc.

The signs + and — , besides indicating the operations of
addition and subtraction, are also used in Algebra to distin-
guish between the opposite states of magnitudes like the
above.

Thus, in financial transactions, we may indicate gains or
assets by the sign -t- , and losses or debts by the sign — ; for
example, the statement that a man's property is — $ 100,
means that he has debts or liabilities to the amount of $ 100.

Again, in the thermometer, we may indicate temperatures
above zero by the sign +, and temperatures beloto zero by
the sign — ; for example, + 25° means 25° above zero, and
— 10° means 10° below zero.

Also, we may indicate north latitude and ivest longitude
by the sign +, and south latitude and east longitude by the
sign — ; thus, a place in latitude — 30°, longitude + 95°,
would be in latitude 30° south of the equator, and in longi-
tude 95° west of Greenwich.

EXERCISES.

11. 1. At 7 A.M. the temperature is — 13"; at noon it is
8° warmer, and at 6 p.m. it is 5° colder than at noon. Re-
quired the temperatures at noon and at 6 p.m.

2. At 7 a.m. the temperature is +6°; at noon it is 14"
colder, and at 6 p.m. it is 8° warmer than at noon. E-equired
the tempeT^atures at noon and at 6 p.m.

10 ALGEBRA.

3. What is the difference in latitude between two places
whose latitudes are + 67° and - 48° ?

V 4. A man has bills receivable to the amount of $ 480, and
bills payable to the amount of f 925 ; how much is he
worth ?

5. A vessel sails from the equator due north 28°, and
then due south 57° ; what is her latitude at the end of the
voyage ?

6. At 7 A.M. the temperature is — 7°, and at noon -f- 9°.
How many degrees warmer is it at noon than at 7 a.m. ?

7. What is the difference in longitude between two places
whose longitudes are + 29° and — 86° ?

8. The temperature at 6 a.m. is +14°; and during the
morning it grows colder at the rate of 4° an hour. Required
the temperatures at 9 a.m., at 10 a.m., and at noon.

12. Positive and Negative Numbers.

If the positive and negative states of any concrete mag-
nitude be expressed without reference to the unit, the results
are cdXledi positive and negative numbers, respectively.

Thus, in + $ 5 and — $ 3, + 5 is a positive number, and
— 3 is a negative number.

For this reason the sign(+ is called th.Q positive sign, and
the sign -r the negative signT

If no sign is expressed, the number is understood to be
positive ; thus, 5 is the same as + 5.

The negative sign can never be omitted before a negative
number.

13. The Absolute Value of a number is the number taken
independently of the sign affecting it.

Thus, the absolute value of — 3 is 3.

^ POSITIVE AND NEGATIVE NUMBERS. H

ADDITION OF POSITIVE AND NEGATIVE NUMBERS.

14. Tlie result of Addition is called the Sum.

We shall retain for Addition in Algebra its arithmetical
meaning, so long as the numbers to be added are positive.

We may then attach any meaning we please to addition
involving other forms of number, provided the new meaning
is not inconsistent with principles which have been pre-
viously established.

15. If a man gains $ 6, and then loses $ 3, he will be
worth $ 2.

If he owes $ 5, and then gains $ 3, he will be in debt
to the amount of $2.

If he owes $5, and then incurs a debt of $3, he will
be in debt to the amount of $8.

Now with the notation of § 10, losing $ 3, or incurring
a debt of $ 3, may be regarded as adding — $ 3 to his
property.

Whence, the sum of + $ 5 and - $ 3 is + $2 j
the sum of -|5 and +$3 is -f 2;
and the sum of — $ 5 and — $ 3 is — $ 8.

Or, omitting reference to the tmit,

(+5) + (-3) = +2;
(_5) + (+3) = -2;
(-5)+(-3)=-8.

We then have the following rules :

To add a positive and a negative number, subtract the less
absolute value (§ 13) from the greater, and prefix to the result
the sign of the number having the greater absolute value.

To add two negative numbers, add their absolute values, and
prefix a negative sign to the result.

12 ALGEBRA.

16. 1. Find the sum of + 10 and - 3.

Subtracting 3 from 10, the result is 7.
Whence, (+10) + (-3)= + 7, Ans.

2. Find the sum of — 12 and + 6.
Subtracting 6 from 12, the result is 6.
Whence, (-12) + (+6)=-6, Ans.

3. Add - 9 and - 5.
The sum of 9 and 5 is 14.
Whence, ( - 9) + ( - 5) = - 14, Ans.

EXAMPLES.
Find the values of the following :

4. (-7) + (-5). lO- (-61) + (+28).

5. (+9) + (-4). 11. (-!) + (+!

6. (-8) + (+2). j2 }_S\ + }_1'-

7. (+6) + (-15). 13. ^+i%) + (-W).

8. (-11) + (-16). 14. (_i8_7_) + (+i24).

9. (+62) + (-37). 15. (+20tV) + (-13A)'

MULTIPLICATION OF POSITIVE AND NEGATIVE
NUMBERS.

17. The terms Multiplicand, Multiplier, and Product have
the same meaning in Algebra as in Arithmetic.

We shall retain for Multiplication, in Algebra, its arith-
metical meaning, so long as the multiplier is a positive
number.

That is, to multiply a number by a positive integer is to
add the first number as many times as there are units in
the second.

POSITIVE AND NEGATIVE NUMBERS. 13

For example, to multiply — 4 by 3, we add — 4 three
times.

That is, (-4) X (+3) = (- 4) + (- 4) +(- 4) = - 12.

We may then attach any meaning we please to multipli-
cation by a negative number.

18. In Arithmetic, the product of two numbers is the
same in whatever order they are taken.

Thus, 3x5 and 5x3 are each equal to 15.

If we assume this law to hold universally, we have

(+3)x(-4) = (-4)x(+3).

But by § 17, (- 4) X (+ 3) = - 12 = - (3 X 4).

Whence, (+ 3) x (- 4) = - (3 x 4). (§ 9, 4)

We then have the following definition :

To multiply a number by a negative number is to multiply
it by the absolute value (§ 13) of the multiplier, and change
the sign of the result.

Thus, to multiply + 4 by — 3, we multiply + 4 by +3,
giving + 12, and change the sign of the result.

That is, (+4) X (-3) = -12.

Again, to multiply — 4 by — 3, we multiply — 4 by
+ 3, giving — 12 (§ 17), and change the sign of the result.

That is, (-4) X (-3)= + 12.

19. From §§17 and 18 we derive the following rule:

To mtiltiply one number by another, multiply together their
absolute values.

Make the product plus wheri the multiplicand and multiplier
are of like sign, and minus when they are of unlike sign.

14 ALGEBRA.

1. Multiply + 8 by - 5.

By the rule, (+ 8) x (- 5) = - (8 x 5) = - 40, Ans.

2. Multiply - 7 by - 9.

By the rule, (_ 7) x (- 9) = + (7 x9) = +63, Ans.

EXAMPLES.
Find the values of the following :

3. (+ 6) X (- 3). 10. (- 24) X (- 18).

4. (_10)x(+5). 11. ^+1^x^-1).

12. (-^]x -^

6. (-12) X (+4). V IV V 14

7. (-8)x(-8). 13. f-mxf+f^

^ 55J \ 2i

8. (-15)x(+9). 14. (+9|)x(-2f).

9. (+ 11) X (- 16), 15. (- li|) X (- 1^)

ADDITION. 15

III. ADDITION AND SUBTRACTION OF
ALGEBRAIC EXPRESSIONS.

DEFINITIONS.

20. A Monomial, or Term, is an expression (§ 8) whose
parts are not separated by the signs -f- or — ; as 2 ar, — 3 ah,
or 5.

2 ic^, — 3 ah, and + 5 are called the terms of the expression

2 a;2 - 3 a6 + 5.

A Positive Term is one preceded by a + sign ; as + 5 a.
If no sign is expressed, the term is understood to be posi-
tive.

A Negative Term is one preceded by a — sign ; as — 3 ah.
The — sign can never be omitted before a negative term.

21. If two or more numbers are multiplied together, each
of them, or the product of any number of them, is called a
Factor of the product.

Thus, a, h, c, ah, ac, and he are factors of the product ahc.

22. Any factor of a product is called the Coefficient of

the product of the remaining factors.

Thus, in 2 ah, 2 is the coefficient of ah, 2aoih, aoi2h, etc.

23. If one factor of a product is expressed in numerals,
and the other in letters, the former is called the numerical
coefficient of the latter.

Thus, in 2 ah, 2 is the numerical coefficient of ah.

If no numerical coefficient is expressed, the coefficient 1
is understood ; thus, a is the same as 1 a.

16 ALGEBRA.

24. By § 19, (- 3) X a = - (3 X a) = - 3a.
That is, — 3a is the product of — 3 and a.
Then, — 3 is the numerical coefficient of a in — 3 a.
Thus, in a negative term, the numerical coefficient includes

the sign.

25. Similar or Like Terms are those which do not differ
at all, or else differ only in their numerical coefficients ; as
2 a?y and — 7 x-y.

Dissimilar or Unlike Terms are those which are not simi-
lar J as 3 x^y and 3 xy^.

ADDITION OF MONOMIALS.

26. The sum of a and 6 is a + & (§ 3) ; and the sum of a
and — 6 is expressed a + (— &).

27. Required the sum of a and — h.

By § 10, if a man incurs a debt of $ 4, we may regard
the transaction either as adding — f 4 to his property, or as
subtracting $ 4 from it.

That is, adding a negative number is equivalent to subtract-
ing a positive member of the same absolute value (§ 13).

Thus, the sum of a and — 5 is obtained by subtracting
b from a.

Or, o + (— 6)=a — &.

28. It follows from §§26 and 27 that the addition of
monomials is effected by uniting them with their respective
signs.

Thus, the sum of a, —b,c,— d, and — e is

a — b-\-c — d — e.

It is immaterial in Avhat order the terms are united, pro-
vided each has its proper sign.

ADDITION. 17

Hence, the above result may also be expressed
c + a — e — d — b,
— d — b + c — e + a, etc.

29. If the same number be both added to, and sub-
tracted from another, the value of the latter will not be
changed (§9).

That is, a + & — 6 = a.

Hence, terms of equal absolute value, but opposite sign,
in an expression, neutralize each other, or cancel.

30. To multiply 4 by 5 + 3, we multiply 4 by 5, and then
4 by 3, and add the second result to the first.

In like manner, to multiply a by 6 + c, we multiply a by
J>, and then a by c, and add the second result to the first.

That is, a{b -\- c) = ab -\- ac.

31. Addition of Similar Terms (§ 25).

1. Find the sum of 5 a and 3 a. -

We have, 5 a + 3 a = (5 + 3)a (§ 30)

= 8 a, Ans.

2. Find the sum of — 5 a and — S a.

We have, (- 5a) + (- 3a) = (- 5) x a + (- 3) x ct (§19)

= [(-5) + (-3)]xa (§30)

= (-8)xa (§15)

= -8 a, Ans. (§19)

3. Find the sum of 5 a and —3 a.

We have, 5a +(- 3)a =[5 +(- 3)] x a (§30)

= 2 o, Ans. (§ 15)

18 ALGEBRA.

4. Find the sum of — 5 a and 3 a.

We have, (- 5)a + 3a =[(- 6)+ 3] x a (§30)

= (-2)xa (§15)

= — 2 a, Ans.

Therefore, to add two similar terms, find the sum of their
numerical coefficients (§§ 15, 24), and affix to the result the
common letters.

EXAMPLES.

Add the following :

5. 5 a and — 12 a. 9. — be and 6 be.

6. — 7 7n and —8 m. 10. xyz and — 9 xyz.

7. 15 a; and -11 a;. 11. -18mV and -27mV.

8. -lOa^ and 4a2. 12. 36a'6c« and - 19 a^&c*.^

13. Kequired the sum of 2 a, — a, 3 a, — 12 a, and 6 a.

Since the order of the terms is immaterial (§ 28), we may add the
positive terms first, and then the negative terms, and finally combine
these two results.

The sum of 2 a, 3 a, and 6 a is 11 a.

The sura of — a and — 12 a is — 13 a.

Then the required sum is 11 o + (— 13 a), or — 2 a, Ans.

Add the following :

14. 9 a, — 7 a, and 8 a. 15. 13 x, —x, — 10 x, and 5 x.

16. 12 abc, abc, — 6 abc, and — 17 abc.

17. 15 m^, — 11 m^, — 4 m^, m^, and 14 ml

18. 21 a^y*, - 16 xY, - ^y*, 3 x^y*, and - 19 a^y*.

If the terms are not all similar, we may combine the
similar terms, and unite the others with their respective
signs (§ 28).

ADDITION. 19

19. Kequired the sum of 12 a, —5x, —3 2/^, —5 a, 8 a;,
and —3x.

The sum of 12 a and — 5 a is 7 a.

The sum of - 5x, 8x, and - 3 x is 0 (§ 29).

Then the required sum is 7 a — 3 y', Ans.

Add the following:

20. Sab, — 7 cd, — 5 ab, and 3 cd.

21. 6a;, —10z,2y,4:Z, —9y, and — a;.

22. 12 m^, — 2 7/1, — 8 w, 5, — 3 w, — 7 m^, and 11 n.

23. 10 a, — 6 d, —5 c, 12 b, — a, c, — 3 c, and —9 a.

24. 7 X, —4:y, — 3z,9y, —2x, —8x, — 5z, 6y, and — z.

DEFINITIONS.

32. A Polynomial is an algebraic expression consisting of
more than one term ; as a -{- b, ov 2 x^ — 3 xy — 5 y'\

A Binomial is a polynomial of two terms ; as a + &.
A Trinomial is a polynomial of three terms.

33. A polynomial is said to be arranged according to the
•descending powers of any letter, when the term containing
the highest power of that letter is placed first, that having
the next lower immediately after, and so on. Thus,

x*-\-32ify-2a^y^ + 3xf-4:y*

is arranged according to the descending powers of x.

Note. The term — 4 ?/*, which does not involve x at all, is regarded
as containing the lowest power of x in the above expression.

A polynomial is said to be arranged according to the
ascending powers of any letter, when the term containing
the lowest power of that letter is placed first, that having
the next higher immediately after, and so on. Thus,

X* + 3x'y - 2 x^y^ + 3xf - 4:y*

is arranged according to the ascending powers of y.

20 ALGEBRA.

ADDITION OF POLYNOMIALS.

34. A polynomial may be regarded as the sum of its
separate monomial terms (§ 28).

Thus, 2a — 3& + 4cis the sum of 2 a, —3b, and 4 c.

Hence, the addition of polynomials may he effected by unitiny
their terms with their respective signs.

1. Kequired the sum of Qa — 7a^, 3a^ — 2a-\-3y^, and
2 ic^ -- a — mn.

It is convenient in practice to set the expressions down one under-
neatli the other, similar terms being in the same vertical column.

"We then add the terms in each column, and unite the results vyith
their respective signs. Thus,

6 a - 7 a;2
■-2a + 3x2 + 3y8
— a + 2x^ — mn

Sa — 2x^ + Sy^ — mn, Ans.

EXAMPLES.
Add the following ;

2. 3. 4.

7a-5b — 8m2+ 5n^ —19ab- led

- 9 a + 2 & 12 m^ - 16 w^ 8 a6 - 17 cd

3a- & _ 6m2 + 14ri3 5 ab + 13 cd

5. 4a-66 + 3c and 5a + 2&-9c.

6. m^ + 2mw + n^, m^ — 2mn-\-n^, and 2m^— 2w^.

7. 3x — ^y, 7y — 6z, and 5z — 2x.

8. 2a^-5ab~b% 7a^+3ab-9b% and - Aa^-Bab + 8b^

9. 4:X-3x'-ll+5a^, 12 a^ -7 -Sx'-15x,

and U + 6a^-{-10x-9x^.

ADDITION. 21

Note. It is convenient to arrange the first expression in descending

powers of x (§ 33), as follows :

5x3-3x2 + 4x- 11;

and then write the other expressions underneath the first, similar
terms being in the same vertical column.

10. 2a-3b-5c, 8b-\-6c-\-7d, -4a-3c + 2d,

and 7 a — b — 9d.

11. a^ — 3 xy^ — 2 cc^y, 3 x-y — 5 ?/^ — 4 xy'^, 5 xy- — 6 y^ — 7 a^,

and 8y^ + 7 x^ — 9 x-y.

12. 6 a - 8 6 - 2 c, 12c + 9d - 7 a, lib -10 c - 5d,

and —3b — 4:d-\-a.

13. 15a^-2-9a''-3a, 13a - ba' - 6 -7 a%

8 + 4 a - 8 a^ - 7 a2, and IGa^ + 3a^ -lOa-2.

14. 9 a^ -13b'- IS c", IBc' +12b- -8d%

19d'-Ua'-\-3c', and -2b' -16 d' -^11 a".

15. 12a^-a^-{-4:ax'-5a% 18x^ -2a'x-3a^ -13ax^,

15 a'x - 11 a^ - 17 a^ + 3 ax",
and 6 aa;2 _ 8 a^a; - 7 a^ + 9 a^

16. 13a:2^3_43._^g3^^ _9a; + 5 + 16a^ + aj2,

-15-6a:2_7^^]^l^^

and - lOa^ - 12aj + 14a;2 - 17.

SUBTRACTION.

35. Subtraction, in Algebra, is the process of finding one
of two numbers, when their sum and the other number are
given.

The Minuend is the sum of the numbers.

The Subtrahend is the given number.

The Remainder is the required number.

22 ALGEBRA.

36. The remainder when 6 is subtracted from a is ex-
pressed a — b (§ 3) ; and the remainder when — 6 is sub-
tracted from a is expressed a— (—6).

37. Let it be required to subtract — b from a.

By § 35, the sum of the remainder and the subtrahend
is equal to the minuend.

Therefore, the required remainder must be such an ex-
pression that, when it is added to — b, the result shall
equal a.

Now if a + 6 be added to — b, the result is a.

Hence, the required remainder is a + 6.

That is, a — {—b)=a + b.

38. From §§ 36 and 37, we have the following rule:

. To subtract one number from another, change the sign of
the subtrahend, and add the result to the minuend.

SUBTRACTION OF MONOMIALS.

39. 1. Subtract 5 a from 2 a.

It is convenient to place the subtrahend under the minuend.
We then change the sign of the subtrahend, giving — 6 a, and add
the result to the minuend. Thus,

2a

— 5a

— 3 a, Ans.

2.

Subtract

— 5a from —2a.

The student should perform mentally the operation of changing the
sign of the subtrahend ; thus, in Ex. 2, we mentally change — 6 a to
5 a, and then add 5 a to — 2 a.

-2a

— 5a

8 a, Ans.

SUBTRACTION. 23

EXAMPLES.

Subtract the following :

3. 7 from 4. 6.-9 from -25. 9.-5 from 16.

4. 4 from -11. 7. 18 from 5. 10. 12 from -17.~

5. - 15 from -9.- 8. -26 from - 18. 11. -14 from 13.

12. 13. 14. 15. 16.

15a -12a^ -Jab Um^n 27 ayz

6a -31 or' 17 a6 - 8 m^n 34.xyz

17. —xy from xy. 21. — 45 aa?"* from —19 ax*.

18. - 16 a^ from - 44 a'. 22. 31 a'b' from 8 a'-ft^.

19. 21 m'n^ from 39 mV. 23. From 8 a take - 12 b.

20. 19a6c from -6a6c. 24. From - 3 m^ take 4 7i2.

25. From — 23 a take the sum of 19 a and — 5 a.

Note. A convenient way of performing examples like tlie above
is to write the given expressions in a vertical column, change the sign
of each expression which is to be subtracted, and then add the results.

26. From the sum of —18xy and 11 xy, take the sum
of — 29 xy and 17 xy.

27. From the sum of 26 a^ and — 7 a^, take the sum of
— 15 a^ and 48 a^.

28. From the sum of 33 nH and — 16 7i\ take the sum
of 49 n\ — 27 n\ and — 39 n^x.

SUBTRACTION OF POLYNOMIALS.

40. A polynomial may be regarded as the sum of its sep-
arate monomial terms (§ 28) ; hence,

To subtract one polynomial from another, change the sign
of each term of the subtrahend, and add the result to the
minuend.

24 ALGEBRA.

1. Subtract 7ab^-9 arb + %W from 5 a^ - 2 a^6 + 4 ab\

It is convenient to place the subtraliend under the minuend so that
similar terms shall be in the same vertical column.

We then mentally change the sign of each term of the subtrahend,
and add the result to the minuend. Thus,

5 a3 - 2 cfib + 4 ah"-

- 9 d^b + 7 a62 + 8 63

5 aS + 7 0^6 - 3 ab'^ - 8 b\ Ans.

EXAMPLES.
Subtract the following :

2. 12a? -^ a -1 3. 2a&-|- 5&c-3ca

8 a^ - 6 a + 13 - a6 + 11 6c - 4 ca

4. From x^ — 2 xy + if subtract oi? -\-2xy + y^.

5. From 5a — 35 + 4c subtract 5a + 36 — 4c.

6. From 4r'-9ar+ll.r-18 take 3a^-8ar^+17a;-25.

7. From 8a; — 3^/ — 4z take —z-\-llx — <dy.

8. Take 76-9c-2d from 6 a - 5 & + 12 c.

9. Take 12 a^ + 4 a - 9 from 3 a^ + 8 a^ - 6.

10. Subtract a-^- 7 - 2 x - 6 a- from 5 x-'- 12 + 9 x3-2 ic.

(See Note to Ex. 9, page 21.)

11. Subtract 1 + a? — a — o? from 3 a — 3 a^ + 1 — a*.

12. Take 81 6^ + 4a2- 36a6 from -30a6 +9a- + 25 fe^.

13. From 10 ar*- 21 x-^- 11 a; take - 15ar - 20a: + 12.

14. From 17 o? - 12 ah^ + 5 6^ take 8 a^ - 3 a^ft + 13 h\

15. Take —3?-\-3a?y—^xy'^+'if' from ar'- 2a^2/— 2a^+2/^.

16. Take 6c-5d-9 6-4a from -10 6-2c+3a-9d

17. Subtract 4 - 3 x - ar + 8 a.-^ + 10 x*

from 9-7x + 6ar-12x3 + 5x*.

SUBTRACTION. 25

18. Subtract 2x2 -a;?/ + 8 2/' -9 a; -14 2/

19. From 7a-lla'-8 + 6a'

subtract 16 a^ — 9 + 2 a^ + 15 a — 10 a*.

20. From x^ -\-3x^y — a^y^ + 5 x^y^ — 4 xy*

subtract 8 x*y — 7 x^y- — 6 xry^ + 11 xy* — y^.

21. From a^+2ab-{-b^ subtract the sum of —a^-\-2ab — b^

and - 2 a^ + 2 b'.

Note. Write the expressions one underneath the other, similar
terms being in the same vertical column, change the signs of the terms
of each expression which is to be subtracted, and add the results.

22. From the sum of 3 0"+ 2ab-b^ and 5 a^- 8 a6 + 6 b^,

take 6 a^ — 5 ab -\- 5 b^.

23. Subtract the sum of 9aj^ — 8a; + a^ and 5— a^^ + aj

from 6 a;^ — 7 a; — 4.

24. Subtract the sum of x + y — 8z and — 4 a; + 9 y from

the sum of 9x~2y — z and —5x + 6y — 7z.

25. Take the sum of 6 — 4 a^ — x and 5 a; — 1 — 2 a;^ from

the sum of 2a^ + 7 — 4a; — 5a^
and 3x^~6x^-2 + 8x.

26. From the sum of 2a-36 + 4d and 26 + 4c — 3d, take

the sum of — 4a — 46 + 3c — 2d and 3a — 2c.

27. From the sum of 9a^ — a^ — 5 and 3 a^ — a + 1, take

the sum of -8aH13a + 3 and 5a'+2a^-6a.

26 ALGEBRA.

IV. PARENTHESES.

REMOVAL OF PARENTHESES.

41. The expressions a — b-\-(c — d)

and a~b — (c — d)

indicate that the expression c — d is to be respectively
added to, and subtracted from, a— b.

If the operations be performed, we have by §§ 34 and 40,
a — & + (c — d) = a — & + C — d,
and a — b — (G — d) = a — b — c + d.

In the first case, the signs of the terms within the paren-
thesis are not changed when the parenthesis is removed;
while in the second case, the sign of each term within is
changed, from + to — , or from — to +.

We then have the following rules :

A parenthesis preceded by a -\- sign may be removed without
changing the signs of the terms enclosed.

A parenthesis preceded by a — sign may be removed if the
sign of each term enclosed be changed, from + to —,or from
— to +.

42. The above rules apply equally to the removal of the

brackets, braces, or vinculum (§ 7).

It should be noticed in the case of the latter that the
sign apparently prefixed to the first term underneath is
in reality prefixed to the vinculum ; thus, •\- a — b means
the same as + (a — b), and —a — b the same as — (a — b).

43. 1. Remove the parentheses from

2a -36 -(5a'- 46) + (4a -6).
By the rules of § 41, the expression hecomes

2o — 36-5a + 46 + 4a-6 = a, Ans.

PARENTHESES.

27

Parentheses are often found enclosing others; in this
case they may be removed in succession by the rules of
§ 41 ; and it is better to remove first the innermost pair.

2. Simplify 4 a? -{3a; + (—2 a;- ^^^)}.

Removing the vinculum first, and the others in succession, we have

4aj-{3a; + (-2a;-a;-a)}
= 4x-{3a; + (-2x-a; + a)}
fe4a;— {3a;^2a5 — « + a}
ss4:X-Sx +2z + x-a = 'ix- a, Ans.

EXAMPLES.

Simplify the following expressions by removing the
parentheses, etc., and uniting similar terms:

3. 8a + (5&-a)-(-76 + 2a).

4. 4m-[2m + 9n] — |-5m — 6w|.

5. x-\-y — z-{-y — z — x — z — x + y.

6. ah -U^ - {2o? -h-^ -\-5a? + 2ah -^h%

7. m^ — 3 mw + 5 m^ — mn — 6 ri^ — [8 m^ — 4 mw — 7 n^].

8. 4a;- (5a;- [3a;-l]).

9. a — {h — c + d-\-e).

10. 5a6-[(3afit-10)-(-4a6-7)].

11. 7ci?+{-Sa? + 2x-'B)-{4.a?-Qx-2).

12. m — (6m — 7w) — | — 3w + 4w — (2m-3w)^

13. 17 -[45 -(9 -23 -32)].

14. 3a-(5a-{-7a + [9a-4]|).

15. a; - [2 a; - (- a; + 1) + 3] - \Qx - [_ (a; _ 3) - x] \,

16. x-{y + z-\x-{-x-y)-\-z'])-\-\z- 2x~y\.

17. 2w-[3n-|47i-^r=^^-(-5w-9)].

18. 28 - { - 16 - (- 4 +[55 - 31 + 47])|.

28 ALGEBRA.

19. a-(2a-[3a-{4a-5a-l|]).

20. c-l2c-(6a-b)-lc-oa+2b-(-5a+6a-3b)\'^

21. x-[y-lx- z-x-y + z\ + (2 x-\-^ + Vl)]-

22. o X - [2 X - (- X- 12 X- ^r=^| - 3 .r) - 3 x"].

23. a — I — a — [— a — (— a — \ — a — a — ll)]\.

INTRODUCTION OF PARENTHESES.

44. To enclose any number of terms in a parenthesis, we
take the converse of the rules of § 41 :

Any number of terms may be eyidosed in a parenthesis
preceded by a + sign, without changing their signs.

Any number of terms may be enclosed in a parenthesis pre-
ceded by a — sign, if the sign of each term be changed, from
4- to —, or from — to +.

1. Enclose the last three terms of a — & + c — d + e in
a parenthesis preceded by a — sign.

Result, a— b—(— c + d — e).

EXAMPLES.

In each of the following expressions, enclose the last
three terms in a parenthesis preceded by a — sign:

2. a + b — c-d. 6. x^-{-f + ^-3xyz.

3. 3f — 5x^ — Sx-\-7. 7. a — b — c + d + e.

4. m^ + m-n + mn^ + n^. 8. a* + 6a^ + a- - 9a + 2,

5. a- — b- -\- 2 be — c^. 9. a~ — m- — 2 mn — ji^.

10. In each of the above results, enclose the last two
terms in parenthesis in brackets preceded by a — sign.

MULTIPLICATIOX. 29

V. MULTIPLICATION.

45. The Law of Signs.

If a and b are any two numbers, we have by § 19,

{+ a) X {+ b) = -{- ab, {+ a)x(- b) = - ab,

{— a) X {-\- b) = — ab, (— a)x{—b) = + ab.

From these results, we may state the Rule of Signs in
Multiplication as follows :

+ multiplied by +, and — multiplied by —, produce +;
+ multiplied by —, and — multiplied by +, produce — .

Or, as it is usually expressed with regard to the product
of two terms.

Like signs produce +, and unlike signs produce — .

46. The Index Law.

Let it be required to multiply a^ by a^.

By § 6, a^= a X a X a,

and a^ = a X a.

Whence, a^xa^ = axaxaxaxa=a^.

Therefore, the exponent of a letter in the product is equal
to its exponent in the multiplicand plus its exponent in the
multiplier.

Or in general, if m and n are any two positive integers,

a" X a

n ^m-j-n

A similar result holds for the product of three or more
powers of a.

Thus, a^ X a* X a^ = a^+*+^ = a^.

30 ALGEBRA.

MULTIPLICATION OF MONOMIALS.

47. Let it be required to multiply 7 a by —2 b.

We have, - 2 & = (- 2) x &. (§ 45)

Whence, 7a x (-25) = 7a x (-2) x &.

Then since the order of the factors is immaterial (§ 18),

7ax(-2&) = 7x(-2)xax6

= -Uab. (§19)

48. From § § 45, 46, and 47, we derive the following rule
for the multiplication of two monomials :

To the product of the absolute values of the numerical coeffi-
cients, annex the letters; giving to each an exponent equal to its
exponent in the midtiplicand plus its exponent in the rmdtiplier.

Make the product + when the midtiplicand and multiplier
are of like sign, and — tvhe7i they are of unlike sign.

1. Multiply 2 a' by 9 a*.

By the rule, 2a^x9a* = 2x9x a^* = 18 a^, Ans.

2. Multiply a^b'-c by - 5a^bd.

We have, a%-c x ( - 5 a'^bd) = - 5 a^b^cd, Ans.

3. Multiply —7 a;'" by 4 ic*.

We have, (— 7x'")x 4x3 = — 28a;'"+3, -4ns.

4. Multiply -3 a;" by —8 a;".

We have, ( - 3 x") x ( - 8 x») = 24 x»+" = 24 x^", Ans.

EXAMPLES.

Multiply the following :

5. 7 a^ by 3 a^ 7. 5 xyz by — 11 xyz.

6. -14a6 by 2cd. 8. -15a«6 by -4a6'.

MULTIPLICATION. 31

9. -9mV by 7m-^i^. 13. - a^i^c" by - a&V.

10. -6a^b^ by - 6V. 14. -8*"*?/" by 12 x"^".

11. Sx'z' by -82/«2^ 15. lOa^^V by 9a'c'd^

12. 12 a^ftc by 6bccP. 16. 16 a^?/' by -Sa^^p^'".

49. We have by § 45,

(— a)x(— b)x(— c) = (ab) x(— c)=— a&c; (1)
(— a) X (— &) X ( — c) X (— c?) = (— abc) x (— d), by (1),

= abed ; etc.

That is, the product of three negative terms is negative ;
the product of four negative terms is positive ; and so on.

Hence, the X)roduct of any mimber of terms is positive or
negative according as the number of negative terms is even or
odd.

1. Eequired the product of —2a^b^, 6bc^, and — 7 c^d.

Since there are two negative terms, the product is positive.

Whence, ( - 2 a-b^) x (6 bc^) x ( - 7 c^d) = 84 a-b*c~d, Ans.

EXAMPLES.

Multiply the following :

2. 3 a^, 5 a^, and — 6 a*.

3. -Ax", -9tf, and 2z^

4. cc^'"?/", y"^, and a^z''.

5. -12a^6^ -6V, and -8(^a*.

6. a', 3 a, 5 a', and — 7 a^.

.7.-2 a^i^c, 2 a'bd^, - 2 ac^d^ and 2 6Vd.

8. 2a^/, -3?/2^ -42^0;*, and -Bx^z^.

9. — a"*.^^, — &""?/, — a^"'?/', and — b"x.

10. 5a6*, -4aV, -a^d^Gfe^c, and -Sc^d^

32 ALGEBRA.

MULTIPLICATION OF POLYNOMIALS BY MONOMIALS.

50. In § 30, we showed that the product of a + b and c
was ac + be.

We then have the following rule for the product of a
polynomial by a monomial :

Multiply each term of the polynomial by the monomial, and
unite the results with their proper signs.

1. Multiply -8a^hj2x'-5x + 7.
Multiplying each term of 2x2 — 5x + 7 by — 8x^, vfe have
(2x2-6x + 7)x(-8x3) = - 16x5 + 40x*-66x3, Ans.

EXAMPLES.
Multiply the following :

2. 4 a — 9 by 5 a. 7. m^w^ by m^ — 2 mn + w^.

3. 8x^y-5xf hj - 3 xy\ 8. 8 a^ft" - 9 ab^ by - 6 a'b^

4. a^ - ab + b'" hj ab. 9. 6a^-5x^-7 x' by -8x^.

5. 3 ^2 + a; - 8 by - 9 a^. (See note to Ex. 9, p. 21.)

6. ~7a^ hj 2a^-6a^-7. 10. -4&"-a"+5a6 by 4a2&3.

11. — x^y^ by cc^ — 3 x^y + 3 xy^ — y^.

12. 5a3 + 9-8a*-4a-a2 by 7a^

13. — 2 mn by 3 m^ — 6 m^n — 7 mn^ -\-2n^.

MULTIPLICATION OF POLYNOMIALS BY POLYNOMIALS.

51. Let it be required to multiply a + b hj c -\- d.

As in § 30, we multiply a + b by c, and then a + 6 by c?,
and add the second result to the first.

That is, {a -\- b){c + d) = (a + b)c + (a -\- b)d
= ac + be + ad + bd.

MULTTPLTCATION. 33

We then have the following rule :

Multiply the mxdtiplicand by each term of the multiplier,
and add the partial products.

52. 1. Multiply 3a-4& by 2a-5 6.

lu accordance with the rule, we multiply 3 a — 4 6 by 2 a, and
then by — 5 6, and add the partial products.

A convenient arrangement of the work is shown below, similar
terms being in the same vertical column.

3a -46
2a -56

Qa- — 8ab

- 15 ab + 20 b^

6 a2 - 23 a6 + 20 b'^, Ans.

Note. The work may be verified by performing the example with
the multiplicand and multiplier interchanged.

2. Multiply 4:ax^-\-a^ — Scc^-2a^xhj2x-\-a.

It is convenient to arrange the multiplicand and multiplier in the
same order of powers of some common letter (§ 33) , and write the
partial products in the same order.

Arranging the expressions according to the descending powers of
a, we have

a3-2a2x + 4ax2 -8x^
a +2x

a* - 2 a^x + 4 a^x"^ - 8 ax^

2a^x-4 a?x^ + 2,ax^-lQx^

a* - 16x*, Ans.

EXAMPLES.
Multiply the following :

3. 2a + 5 by 3a + 7. 6. - 7a& + 2 by - 4a6- 6.

4. 5 a — 8 by 6 a — 1. 1 . x^ — xy -{- y'^ hj x -\- y. ^

5. -4a;-52/by 8a; + 3y. 8. 2a* + 7a- 9 by 5a - 1.

34 ALGEBRA.

9. 3a^- a; + 4 by 4a; -3.

10. -8n-\-5n'-3 by 2 + w.

n. 3a-2b by 9a' + 6ab +4:ly^.

12. a — 6 + c by a — 6 + c.

13. 6 m^ — 5 mn — 8 w^ by 2 m^ + 3 mhi.
U. x' + 3x^-7x-6 by 3x-4..

15. m^ + mw + ri^ by m^ — mn + n^.

16. 8a3-4a2 + 2a-l by 2a + l.

17. dxr'-S + Gxhy 8x^4: + 7x'.

18. 6 w - 8 + 4 ?i2 by - 4 + 2 w^ _ 3 n.

19. 3a2-5a6-862by4a2-9a6-7&2.

20. 2aJ + 6z-4?/ by 2?/-32 + a;.

21. 4a+6& + 10c by 2a-3& + 5c.

22. a^ - 2 a^ + a - 2 by a^ + 2 a + 3.

23. x^ + 2 a;^ + 4 a;2 + 8 a; + 16 by a; - 2.

24. m^ + n^ + mn^ + ni^n by m^n — mv?.

25. -5a;2 + 9 + 2a,-3-4a; by 5a;2_;l^_f_(3a.^

26. 4:a^'^y — 5a^y'"-^ by 2 x^^-y +4 _ 3 ^^_

27. 3m3-5m2 + 4m-l by 2m2-m-3.

28. 16a*-24a^ + 36a2-54a + 81 by 2a + 3.

29. a^-3a'b+3ah'-b^ by a'-2ab + b^

30. a;3_2a.-2-a;-l by af' + 2 a;^ _ ^ _^ ;j^^

31. a^ - 6 a^ + 12 a - 8 by a^ + 6 a^ + 12 a + 8.

32. m^ — 6 mn — In^ by m^ — 2 m'n — 5 mn^ + 4 w^

33. a;3-3+2a;2_3. by 3_a;+a^_2a;2_

34. o? + W+c'-\-ab — bc + ca by a — 6 — c.

35. 6 ar' - 4 a;2^ - 3 ^2/2 + 2 1/^ by 2 a;2 + iC2/- 2 jr*.

MULTIPLICATION. 35

Find the product of the following :

36. x — 2,x — 3, and a; — 4.

37. tt + 5, 2 a — 3, and 4 a - 1.

38. X ~y, XT +xy -\- if, and a^ + 2/^

39. 3w-8, 4n + 7, and 5n-6.

-iO. a — a;, a + x, a^ + x^, and «■* + «*.

41. m — 4 71, 2 m 4- 3 w, and 2 m^ + 5 mn — 12 n^.

42. a + 1, a — 1, a^ + a + 1, and a^ — a + 1.

43. a;' + a; + 1, ar — a; + 1, and x* — ar^ + 1.

44. a + h, a — b, 2a~3b, and 2a + 36.

45. a;+3, 2a; + l, 2a;-l, and 4ar'- 12a^ + a; - 3.

53. 1. Simplify (a-2xy-2(x + 3a){a- x).

To simplify the expression, we should first multiply a — 2x by
itself (§ 6) ; we should then find the product of 2,x + 3a, and a — x,
and subtract the second result from the first.

a —2x
a —2x

Ha

+

X

a

-

X

3a2

+

ax

-

Sax-

a;2

3a2

—

2ax-

x2
2

a^ — 2ax

— 2ax + 4x^

6o2_4ax-2a;2

Subtracting the second result from the first, we have

a2 - 4 ax + 4a;2 - 6a2 + 4 ax + 2 x2 = - 5 a2 + 6x2, Ans.

EXAMPLES.

Simplify the following :

2. (3a;-8)(a; + 6) + (2aj-7)(4a;-f 9).

3. (2a + 5)(3a-7)-(2a-5)(3a + 7).

36 ALGEBRA.

4. (a — VI) (b + n) + (a -\- m) (6 — w).

5. (x — y + zy—{x + y— zf.

6. (a-&-c + cf)2.

7. (2 a; + 3)2(2 a:- 3)1-

8. (a + h) (a^ - 52^ _ (^ _ j)^ (^2 _^ 52^^

9. (3aj-5y)-'-5(a:-2/)(aj-52/).

10. (a + a;) (a^ + :^) [a^ - a; (a - a;)].

11. (a - &) (a^ + a^h + «&== + }f) [(a^ + 6^)2 _ 2d'h''\

12. (a; + 1) (a; + 2) (a; + 3) - (a; - 1) (a; - 2) (a; - 3).

13. {x -y){y-z)- {x-z) (y-z)-(x- y) (x - z).

14. (a+b-]-cy-(a + bf - c(2(i + 26 + c).

15. (a + 1) (2 a + 5) (4 a - 3) + (a - 1) (2 a - 5) (4 a + 3).

16. (a; + 2/ - 2)2 + (2/ + 2 - a;)2 + (2 + .-c - ?/)2.

17. 2 (a + 2 X) (a - 2 a;) [(a + 2 xf + (a - 2 a;)^].

18. (a + & + c)- - (a + 6 - c)2 - (a - 6 + c)2 + (a - 6 - c)'.

19. [(m + n)2 + (m - w)^] [(2 m + 71)^ - (m - 2 jt)^].

20. (a + b - c){b + c - a){c -]- a - b).

21. (a + 6)8-(a-&)3.

22. (a; + 2/ + 2)3-3(ar' + 2/2^2;2)(a; + 2/ + 2).

DIVISION. 37

.VI. DIVISION.

54. Division, in Algebra, is the process of finding one of
two numbers, when their product and the other number
are given.

The Dividend is the product of the numbers.
The Divisor is the given number.
The Quotient is the required number.

55. The Law of Signs.

Since the dividend is the product of the divisor and quo-
tient, the equations of § 45 may be written as follows :

(+ a6) -=- (+ a) = + 6, {- ah) ^ {+ a) =~h,

(- ab) ^ {— a) = + b, (+ a6) h- (- a) = - 6.

From these results, we may state the Rule of Signs in
Division as follows :

+ divided by +, and — divided by — , produce + ;

+ divided by — , and — divided by +, produce — .

Hence, in Division as in Multiplication,

Like signs produce +, and unlike signs produce — .

56. The Index Law.

Let it be required to divide a^ by a^.

The quotient must be a number which, when multiplied
by the divisor, a", will produce the dividend, a^.
Now if a^ be multiplied by a^, the product is a*.

Whence, ^ = a^.

a-

Hence, the exponent of a letter in the quotient is equal to
its exponent in the dividend minus its exponent in the divisor.

88 ALGEBRA.

DIVISION OF MONOMIALS.

57. Let it be required to divide — 14 a% by 7 a^.

We find a number which, when multiplied by 7a^, will
produce — 14 a?h.
That number is evidently — 2 &.

Whence, =^ = -^^-

58. From §§ 55, 56, and 57, we derive the following rule
for the division of two monomials :

To the quotient of the absolute values of the numerical
coefficients, annex the letters; giving to each an exponent
equal to its exponent in the dividend minus its exponent in
the divisor, and omitting any letter having the same exponent
in the dividend and divisor.

Make the quotient + when the dividend and divisor are oj
like sign, and — when they are of unlike sign.

1. Divide 54 a^ by -9a\

By the rule, -^^ = - 6 a'-* = - 6 «», Ans.

^ -9a*

y^

2. Divide - 2 a^'bhd* by abd\

We have, - 2 a^'^cd* ^ _ g ^2^0, Ans.

abd^.

3. Divide - 91 a^'^y^'z' by - 13 aj^i/V.

We have, ~ ^^ ^^'^V^^ - 7 x^r'^-'^gr-Z = 7 xmsr-S^ ^ni.

EXAMPLES.
Divide the following :

4. 35 by -5. 6. -64 by -4.

5. _44 by 11. 7. -84 by 7.

DIVISION. 39

8. -144 by -8. 17. 40 mW by 5mhi.

9. 168 by - 12. 18. - 33 a^xV by - 3 a'y.

10. 16 a^ by 4 a*. 19. - 36 a*»+i by 12 a^—^.

11. -ISa^y by 2xy. 20. 81a^6V by 9 6V.

12. 2m'n' by — mV. 21. GBx^fz^ by -ISxf.

13. - a^ftV by - a'b^c. 22. - a^ft'' by - a«6'.

14. -6a;Y^ by 6afy^. 23. 54x3^" by 9a;"2/".

15. -24a%2by -8aV. 24. 98a96Vby -Ua'bc'.

16. 28.^'^!/23 by -Ta^^l 25. -143m%«_p^ by llrnVj/.

DIVISION OF POLYNOMIALS BY MONOMIALS.

59. We have, a(b + c) = ab + ac.

Since the dividend is the product of the divisor and quo-
tient (§ 54), we may regard ab + ac as the dividend, a as
the divisor, and 6 + c as the quotient.

Whence, 2:t±^=b + c.

a

We then have the following rule :

Divide each term of the dividend by the divisor, and unite
the results with their proper signs.

1. Divide 9 a^ft^ - 6 a*c + 12 a^Jc' by -3al

9 a362 -6a*c + 12 a^bc^ = _3ab^ + 2a^c - i bc% Arts.

EXAMPLES.
Divide the following :

2. 16a;^ + 28a^-24a58 by 4a:».
, 3. 104mw^ — 39m^n by — 13mn.
4. 6 a^iV - 15 a«6V + 3a*b^c by - 3 d'b\

40 ALGEBRA.

5. -63c(^yh^-84:x'y*z' by 7a^yz\

6. 20 m^n'' — 45 mhi' — 35 mW by — 5 mW.

7. - 24a" + 108 a' + 84a^ by 12 a«.

8. 40 a^bc - 24 aft^c - 32 a&c^ by - 8 a6c.

9. 72iK^''-9a^ + 54iB«-99a;* by -9a;^

10. — 2 aj^y + 6 o.-^^/' ~ 6 x^y^ + 2 a;?/* by —2xy.

11. 60 a" - 30 a^ + 15 a}" - 45 a« by 15 a'.

12. a»'"&'' - 3 a2'"62» 4. 2 a^ft^ by a'"6".

13. 48 a^&«c* + 36 a?¥& - 30 a«6V by 6 a^fe^c*.

14. - 88 x'fz" + 55 xyh' + 66 ar^^/^z^ by - 11 xfz\

15. a;"+^2/'"^^2;^ — ayz* — a;"?/»2;'' by — x'^y^z^.

DIVISION OF POLYNOMIALS BY POLYNOMIALS.

60. Let it be required to divide 12 + 10 or' — 11 cc — 21 x'
by 2a;2-4-3x.

Arranging each, expression according to the descending
powers of x (§ 33), we are to find an expression which,
when multiplied by the divisor, 2 ar — 3 a? — 4, will produce
the dividend, 10 ar^ - 21 a.-^ - 11 a; + 12.

It is evident that the term containing the highest power
of x in the product is the product of the terms containing
the highest powers of x in the multiplicand and multiplier.

Therefore, 10 21? is the product of 2 x^ and the term con-
taining the highest power of x in the quotient.

Whence, the term containing the highest power of x in
the quotient is 10 a^ divided by 2 ar, or 5 x.

Multiplying the divisor by 5 a', we have the product
10 a;^ — 15 a:^ — 20 X ; which, when subtracted from the divi-
dend, leaves the remainder — 6 a^ + 9 a; + 12.

This remainder must be the product of the divisor by the
rest of the quotient ; therefore, to obtain the next term of
the quotient, we regard — 6a^ + 9x + 12 as a new dividend.

DIVISION. 41

Dividing the term containing the highest power of f,

— 6 ar^, by the term containing the highest power of x in
the divisor, 2 a^, we obtain — 3 as the second term of the
quotient.

Multiplying the divisor by — 3, we have the product

— 6a^4-9a; + 12; which, when subtracted from the second
dividend, leaves no remainder.

Hence, 5 cc — 3 is the required quotient.

It is customary to arrange the work as follows :

10 a:^ - 21 a^ - 11 a; + 12 I 2a^-3a;-4, Divisor.
10 a^ - 15 ar' - 20a; | 5a; - 3, Quotient.

- 6af+ 9a; + 12

- ex'-i- 9a; + 12

Note. The example might have been solved by arranging the
dividend and divisor according to the ascending powers of x.

From the above example, we derive the following rule :

Arrange the dividend and divisor in the same order 6j
powers of some common letter.

Divide the first term of the dividend by the first term of the
divisor, and write the result as the first term of the quotient.

Multiply the lohole divisor by the first term of the quotient,
and subtract the product from the dividend.

If there be a remainder, regard it as a new dividend, and
proceed as before; arranging the remainder in the same order
of jioivers as the dividend and divisor.

61. 1. Divide9«&2 + a3_9j3_5(j2^by 3 52^^2_2a6.
Arranging according to the descending powers of a,
a3 _ 5 a26 -i- 9 ab^ - 9 &3 l g^ - 2 a& + 3 6^

a^-2 a^b + 3 ab^ | a - 3 6, Ans.

-3a26 + 6a62
-3a2ft + 6a&2_968

Note 1. In the above example, the last term of the second divi-
dend is omitted, as it is merely a repetition of the term directly above.

42 ALGEBRA.

Note 2. The work may be verified by multiplying the quotient

by the divisor, which should of course give the dividend.

2. Divide 8 + 18 x" - 56 a;^ by - 6 a^ + 4 + 8 x.
Arranging according to the ascending powers of x,

4 + 8x-6a;2)8-56a;2 + 18x4(2-4x-3a;2, Ans.
8 + 16 X - 12 x2

-16x -44x2 + 18x*
-16x -32x2 + 24x3

-12x2-24x3+ ISx*
- 12x2- 24x3 + 18x*

EXAMPLES.
Divide the following :

3. Wx'-llx-U by 3a; + 2.

4. 25 m^ + 40 mw + 16 n^ by 5 m + 4 n.

5. 12a2-28a + 15 by 6a -5.

6. K^ - Gar* - 19 a; + 84 by a - 7.

7. 8 m^ + 27^3 by 2 m + 3?i.

8. ar' — 64i/^ by a; — 4?/.

9. 8-16a + 6a2 by 3a -2.

10. 50x2^=^-18 by 3-5a^.

11. 10 a'b^ - 18 a'b* - 3 a'b^ by 2 a^ft - 3 ab\

12. 2 m^ — 8 m^n + 18 mn^ by 2 m^ — 6 mn.

13. 20 + 36a3-49a by 12a2 + 5-16a.

14. 2 a^b^ - 3 a*b^ - 7 a^b* + 4 a^i* by a^b - a^ft" _ 4 ab^.

15. a" - &2 _f- 2 6c - c^ by a + 6 - c.

16. 4.f-16x'y + 6xy^ + 6oi^ by Sa.-^ - 2/2 _ 2a;2/»

17. 39mw=^ + 30m3 - 20w^ - 43m^w by 6m - 5w.

DIVISION. • 43

18. 4:a*- 90^ + 80 a -25 by 2a« + 3a-6.

19. 4:X + x* + 3 by 3-\-x^-2x.

20. n* - 16 by 2n* + 8 + 4 w + n\

21. 6m*-19m3 + 22m + 5 by 3m-5.

22. .r^ + y^ + x^f by y^ + a^ - xy.

23. l-16a« by l+2al

24. 16x^-81?/* by 2a; -3y.

25. - 9m=' - 16 + m* - 24m by 3 m + m'* + 4.

26. 9a^+4-13x2 by 3a^-2 + a;.

27. 2a^-a3 + 8a-5 by 2a2-3a + 5.

28. 13a^ + 71a;-70a^-20-f-6x* by 4 + 30^-70;. .

29. 4mV4-w« + 16m* by 2mn^ + ^m' -{-n*.

30. a.-^ + 32 by a; + 2.

31. 120a* + 26a3-llla2_i4a + 24 by (3a + 2)(4a - 3),

32. (2 ??i2 - m -1)(3 m^-\-m- 2) by (2m + 1)(3 m - 2).

33. a^+243 by 9a2 + 81-3a3-27a + a'.

34. 4 a^ "•+y - 16 x^+y +1 + 12 ar'?/^ "-^ by ^+^y - 3 a;?/»-^

35. 6a« -6a6* by -36 + 3a.

36. a' - a*b - a¥ + ¥ hj a^-2ab+ h\

37. 8m' - 14m2 - 187» + 21 by 4m3 + 6m -7.

38. 16 a" - 96 a? + 216 a? - 216 a + 81 by (2 a - S)\

39. 7ar5-6aT'-28 + 81a;2_^3^_25a;4 by 4-3a^-5a;.

40. 2 a;*' -ea;'- a;'' -9x2 + 3a; -9 by 2a;' -a; + 3.

41. 70a - 50 - a^ - 37a- by 6a - 5 - a^ - 2al

42. af - 81 xy* +2^3 f - 3x'y by 9xy^ + x^ + 27r^ + Zx^y.

43. 14a;''-23a; + 6a;'' + 6a.-2-lla;5^5 _12a;3

by 5x-3x' + 2a^-l.

44 ALGEBRA.

44. 4a«-49a^ + 76a2-16 by 20" + 5a^ - 6a-4:.

45. m^ — 6 vi*rr + 9 m^n* — 4 w^ by (m + w)(wi — n)(m + 2n).

46. 8a2-10a6 + 18ac-3&2 + 86c-5c2by 2a-3b + 5c.

47. a;2p _ 2/2« + 2 ?/'2'' — z-'' by a^^ — ?/' + 2''.

The operation of division may be abridged, in certain
cases by the use of parentheses.

48. Divide ct^ +(a — b + c)x^ + (— a6 — &c + ca)x — abc

by x + a.

x^+(a — b + c)x^+( — ab — bc+ca)x—abc \ x + a

«3 +ax2 \ x^+( — b + c)x—bc, Ans.

(_6+c)x2
, ( — 6 + c)x2 + ( — a6 +ca)x

— bcx

— bcx—abc

Divide the following :

49. a^ +(— a-\-b — c)aP +(— ab — bc + ca)x-^ ahc

by 0^ + (— a + b)x — ab.

50. a^ + (a + & + c)x^ + (a& + 6c + ca)« -f ahc by a; + c.

51. o(?-Jr(2>a~2b - cjx^ +{-&ab + 2bc-3ac)x + &abc

by a;^ + (3 a — c)a; — 3 ac.

52. a(a + 6)a^+(a& + 6- + 6c)a;-c(6 + c) by aa; + (64-c).

53. im,(m — n)a?-^{—mn + n^ — np)x+p{n—p)

by 7;.u;-(n-i9).

54. a:^ + (a — 6 — c) .r^ + (— a6 + 6c — ca) « + abo

by a^ — (6 + c) a; + 6c.

55. .T^ — (a + 6 + c) x^ + (a6 + 6c + ca) a; — a6c by x — a.

56. a^ (6 - c) c^ + a (- 6- + c^ + d") - (6 + c) cZ

by ad — (b + c).

57. a^ 4- (m + n) a — 2 m- + 11 mn — 12n^ by a — m + 4 n.

DIVISION. 45

EXAMPLES FOR REVIEW.

62. 1. Find the numerical value when a = 4, 6 = — 7,
c = — 3, and d = 5, of

^ ^ c4-d

We have, (a + 6)^ =(4 - 7)(4 - 7) = (- 3)(- 3)= 9,

.ad ^-^=--^-^ = ^^-4.

c+d -3+5 2

Then, (a + &)2 -^^:i^= 9 -(- 4)= 9 + 4 = 13, Ans.
c + d

Find the numerical value of each of the following when
a = 5, 6 = — 4, c = — 2, and d = 3 :

2. (a-6)(6 + c)(c-d). 3. 62_ 0^ + 2cd - d^.

4. (a + 36)(4c-ri) + (a-c)(26 + d).

5. a^ - 3 a^ft 4- 3 a&2 _ 63. 8. 3 a'b - 5 b'c + 4. c*d.

g 8ad_6a6^ ^ (« _ ms _ (^ _ d.)3.

oc ca V / \ /

,y a + 26 a-56 jq 26a + 23& + 64c

■4c + d 6c- d' ■ lla + 24 6-7c*

11 2a — 6 3& — c4c — d
6— c c — d d — a

12. Add 9 (a - 6) - 8(& - c), - 3(6 - c) - 7(c - d),

and 4 (c — d) — 5 (a — b).

9(a-6)- 8(5 -c)

- 3(6-c)-7(c-(?)
- 5(a - 6) + 4(c - d)

4(a- 6)- 11(6 -c)-3(c-d), ^ns.

13. Add 4:d'{a + x)-6{b-y), - 3 a2(a + a;) - 2(6 - y),

and — 7 a^ (a + a;) + 8 (6 — ?/).

14. Add 18 (x - yy - 11 (a; + yf, - 9{x-yy -\-7 {x -{- yf,

and — 4 (x — 2/)^ — 5 (x* + 2/)3.

46 ALGEBRA.

15. Subtract 5 (a - 6) - 8 (c + d) from 2 (a - 6) - o{c + d).

16. Multiply 3(x + y)-5 by 3(x + y) + 5.

17. Multiply 7 (a - 5) + 4 by 9 (a - 6) - 8.

18. Divide 6 (??i + yif - (m + n) - 15 by 3 (7n + ?i) - 5.

19. Divide (x-yf-^-l by (» -?/) + !.

20. Add fa + |6-^c and ia-|6 + fc.

21. Add |a-f6 + fc and ^a-h|6-fc.

22. Add ^x-:i^y-^z and - |a; + f 2/ - ^2;.

23. From ^a — f6 + |c take ^a — f6 — |c.

24. Subtract — i^ic + |2/ + |^2 from — |ic + |i/ — |«.

25. Multiply f ic" + i a^ + tf by | a.' - f .

26. Multiply ^a'-^ab+^b' by ^a-ife.

27. Divide ^a;3 + ^^ by |a' + |.

28. Divide ^a' - la'b + ^ab' - ^b^ by ^a-fft.

29. Multiply a'^+^b' - a%^+^ by a^^-^ - b^-\

30. Divide a.-^"-! - a;Y"+2 by .^""-2 + /"+!.

31. Divide a^+'' - ab^^^ by a^+i - b'''-\

32. Add 3(x + 1)=^ -2{x + 1), 5(x' + 1) - 7,

and -(ic + l)2-3(a; + l)+4.

33. From 7 (a; + y)^ - 9 x (x + y) + 4

take 12 (.r + yy + x (x + i/) - 11.

34. Simplify 5x-[Sx-\x-(7 x-8x-4:)\-(9x-5x-2)]

35. Add3^.r=-|a^-J^, _ |i a;^ + J_aj _|,

and ix2-|a;+3^.

36. Multiply ar + (b—c) x — bc by x + a.

DIVISION. 47

37. Divide a""+^b^ - an^"-' by a^+s - &»-♦.

38. Subtract i^ a^ — i « + tV f i^oeq tit ^^ + it ^ ~ ^V

39. Multiply (m -n)' + 2(m - n) + 1

by (m — n)- — 2 (m — n) + 1.

40. Multiply a''" - a"&" + b-" by a^+^fe^ + a&""^*.

41. Simplify (a + bf - 2{a + b){a - b) + (a -- bf.

42. Simplify a-[2a-(6-6c)- [a-(-26-5c)-36-c|]

43. Multiply fa^ _ ^ ^ - | by f a^ _ « _ |. ,

44. Divide |a* - f a^ + ia^ - f by f a^ - a - |.

45. Divide a'-b'-5ab {w" - b^) + 10 a-b' (a - b)

by (a + 6)='-4a6.

46. Divide 12x^+^y''~^ — 13a^'"+*i/''-* — 35ar+^y^-'^

by 4 x^'"+^y"-^ + 5 x'"+ V~^-

47. Multiply (a + 6)a:-2a6 by aj + (a + 6).

48. Divide (a - 6)3 -3(a - 6)^0 + 3(a - &)c2 - c^

by (a-b)-c.

49. Divide x*" + x-^^y^" + ?/*" by ar^" + a;"*^" + 2/^".

50. Multiply fa^-fax-ix^ by fa^ + f aaj + ior'.

51. Multiply a^ -\- (— a + b)x — ab by cc — c.

52. Multiply x^ — .r* + a;*" by x^ — xf + a;*".

53. Divide ^x*-\x^ + ^x-^ by |a^ + ia;-f.

54. Divide a^ -\- (a — b — c)c(^ + {— ab -\- be — ca)x + abc

by a; — c.

55. Simplify (x + y + z) [(x + ?/ + 2)^ — 3 (x?/ 4- yz + za;)].

56. Simplify (a + 6 + c) (— a + 6 + c) (a — 6 + c) (a + 6 — c).

48 ALGEBRA.

VII. SIMPLE EQUATIONS.

63. The First Member of an equation is the expression to
the left of the sign of equality, and the Second Member is
the expression to the right of that sign.

Thus, in the equation 2^ — 3 = 3a; + 5, the first member
is 2 a; — 3, and the second member is 3 a; + 5.

Any term of either member of an equation is called a
term of the equation.

The sides of an equation are its two members.

64. An Identical Equation, or Identity, is one whose
members are equal, whatever values are given to the letters
involved ; as (a + b) (a — b)=o? — W.

65. An equation is said to be satisfied by a set of values
of certain letters involved in it when, on substituting the
value of each letter wherever it occurs, the equation becomes
identical.

Thus, the equation x — y = 5 is satisfied by the set of
values a; = 8, y = 3; for on substituting 8 for x, and 3 for y,
the equation becomes

8 --3 = 5, or 5 = 5;

which is identical.

66. An Equation of Condition is an equation involving
one or more letters, called unknown quayitities, which is not
satisfied by every set of values of these letters.

Thus, the equation a: + 2 = 5 is not satisfied by every
value of X, but only by the value a; = 3.

An equation of condition is usually called an equation.

SIMPLE EQUATIONS. 49

67. If an equation contains but one unknown quantity
any value of the unknown quantity which satisfies the
equation is called a Root of the equation.

Thus, 3 is a root of the equation a; + 2 = 5.
To solve an equation is to find its roots.

68. A Numerical Equation is one in which all the known
numbers are represented by Arabic numerals ; as,

2 ic - 17 = a; ~ 5.

69. A monomial is said to be rational and integral when
it is either a number expressed in Arabic numerals, or a
single letter with unity for its exponent, or the product of
two or more such numbers or letters.

Thus, 3, a, and 2 a^bc^ are rational and integral.

70. If each term of an equation, involving but one un-
known quantity x, is rational and integral, and no term con-
tains a higher power of x than the first, the equation is said
to be of the first degree.

Thus, 3 a; - 5 = 4 ) ^^^ equations of the first degree.

and a^x -[-Ir = c)

A Simple Equation is an equation of the first degree.

PROPERTIES OF EQUATIONS.

71. It follows from § 9, 1 and 3, that :

1. The same number may be added to, or subtracted from,
both members of an equation, without destroying the equality.

2. Both members of an equation may be multiplied, or
divided, by the same number, without destroying the equality.

72. Transposition of Terms.

A term may be transposed from one member of an equation
to the other by changing its sign.

50 ALGEBRA.

Let the equation be x -]- a = b.

Subtracting a from both members (§ 71, 1), we have

x= b — a.

In this case, the term + a has been transposed from the
first member to the second by changing its sign.
Again, consider the equation

X— a = b.

Adding a to both members, we have

x = b -i-a.

In this case, the term - a has been transposed from the
first member to the second by changing its sign.

73. It follows from § 72 that

If the same term occurs in both members of an equation
affected with the same sign, it may be cancelled.

74. The sign of each term of an equation may be changed
without destroying the equality.

Let the equation be a — x=:b — c. (1)

Transposing each term (§ 72), we have

— 64-c=— a + cc.

That is, x-a = c — b;

which is the same as (1) with the sign of each term changed.

SOLUTION OF SIMPLE EQUATIONS.

75. 1. Solve the equation

5a;-7 = 3a; + l.

Transposing 3x to the first member, and - 7 to the second, we
have

6x-3x = 7 + 1.
Uniting similar terms, 2 x = 8.

SIMPLE EQUATIONS. 51

Dividing both members by 2 (§ 71, 2), we have

X = 4, Ans.

From the above example, we derive the following rule :
Transpose the unknown terms to the first member, and the

known terms to the second.

Unite the similar terms, and divide both members by the

coefficient of the unknown quantity.

2. Solve the equation

14-5a;=19 + 3a;.

Transposing, — 6 jc — 3 x = 19 — 14.

Uniting terms, — 8 x = 5.

Dividing by — 8, a; = — -, Ans.

8

Note 1. The result may be verified by putting a; = — - in the
given equation ; thus,

Thatis, 14 + §=19-^-

8 8

Or, i^ = i§Z ; V7hich is identical.

8 8

EXAMPLES.
Solve the following, in each case verifying the answer :

3. 9x = 7x + 28. 10. 7 a; -29 = 16 cc- 17.

4. 8a;-5=-61. 11. 13 - 6a; = 13a; - 6.

5. 6a; + ll = a; + 31. 12. 19 - 16a; = 27 -28a;.

6. 9a;-7^3a;-37. 13. 9a;- 23 = 20a; - i8.

7. 4a;-3 = 8a; + 33. 14. 30 + 17a;= 27a; + 22.

8. 12-13a; = 6-10a;. 15. 24 a; - 11 = 28 + 11a;.

9. 5x + 9 = 14-2a;. 16. 33x + 25 = 41 + 51a;.

17. 14a; + 21-35= -29a; + 44a;-22.

52 ALGEBRA.

18. 32a;-39 = 25x-10a;-141.

19. 12 a; -23 a; 4-55 = 15 a; -75.

20. Solve the equation

(2 X - 1)-' = 2{x + 3) (2 x-3)- 3(6 x - 1).

Expanding (Note 2), 4x2-4a;+l=:4a;2 + 6a;-18-18x-|-3.
Transposing,

4x2_4a;-4x2-6x + 18x= - 18 + 3-1.
Uniting terms, 8 a; = — 16.

Dividing by 8, x = — 2, Ans.

Note 2. To expand an algebraic expression is to perform the
operations indicated.

Solve the following equations :

21. 2(5x + l)-4 = 3(a;-7)-16.

22. 10a; -(3a; + 2)=.-: 9a; -(5a; -4).

23. 8a;-5(4a; + 3)=-3-4(2a;-7).

24. 5a; - 6 (3 - 4a;) = a; -7 (4 + a;).

25. 6a;(3a;-5)-f-141 = 2a;(9a; + l)+13.

26. 19-5a;(4a; + l) = 40-10a;(2a;-l).

27. 2 (4 a; + 7) - 8 (3 a; - 4) = 6 (2 a; + 3) - 7 (2 a; - 3).

28. (5a; + 7)(3a;-8) = (5a; + 4)(3a;-5).

29. (4a; -7)2= (2a; -5) (8a; + 3).

30. (5-3a;)(3 + 4a;) -(7 + 3a;)(l-4a;)= -1.

31. (l-3a;)2-(a; + 5)2 = 4(a; + l)(2a;-3).

32. 6(4-a;)2_5(2a; + 7)(a;-2) = 5- (2a;-|-3)2.

PROBLEMS.

76. For the solution of problems by algebraic methods, no
general rule can be given, as much must depend upon the
skill and ingenuity of the student.

SIMPLE EQUATIONS. 53

The following suggestions will, however, be found of
service :

1. Represent the unknown quantity, or one of the un-
known quantities if there are several, by x.

2. Every problem contains, explicitly or implicitly, pre-
cisely as many distinct statements as there are unknown quan-
tities involved.

All but one of these should be used to express the other
unknown quantities in terms of x.

3. The remaining statement should then be used to form
an equation.

The beginner will find it useful to write out the various
statements of the problem, as shown in Exs. 1 and 2, § 77 ;
after a little practice he will be able to dispense with these
aids to the solution.

77. 1. Divide 45 into two parts such that the less part
shall be one-fourth of the greater.

Here there are tico unknown quantities, the greater part and the
less.

In accordance with the first suggestion of § 76, we will represent the
less part by a;.

The two statements of the problem are, implicitly :

1. The sum of the greater part and the less part is 45.

2. The greater part is 4 times the less part.

In accordance with the second suggestion of § 76, we will use the
second statement to express the greater part in terms of x.

Thus, the greater part will be represented by 4 x.

We now in accordance with the third suggestion of § 76 use the first
statement to form an equation.

Thus, 4 a; + a; = 45.

Uniting terms, 5 x = 45.

Dividing by 5, x = 9, the less part.

"Whence, 4x = 36, the greater part.

54 ALGEBRA.

2. A had twice as much money as B ; but after giving B
$ 35, he had only one-third as much as B. How much had

each at first?

Here there are two unknown quantities : the number of dollars
A had at first, and the number B had at first.

Let X represent the number of dollars B had at first.

The first statement of the problem is :

A had twice as much money as B at first.

Then 2 x will represent the number of dollars A had at first.

The second statement of the problem is, implicitly :

After A gives B $ 35, B has 3 times as much money as A.

Now after giving B $35, A has 2x — 35 dollars, and B a; + 35
dollars ; we then have the equation

x + 35 = 3(2x-35).

Expanding, a; + 35 = 6 a; — 105.

Transposing, — 5 a; = — 140.

Dividing by — 5, x = 28, the number of dollars B had at first ;
and 2 X = 56, the number of dollars A had at first.

Note 1. It must be carefully borne in mind that x can only rep-
resent an abstract number-; thus, in Ex. 2, we do not say, "let x
represent what B had at first," nor "let x represent the sum that-B
had at first," but "let x represent the number of dollars that B had
at first."

3. A is 3 times as old as B, and 8 years ago he was 7
times as old as B. Required their ages at present.

Let X = the number of years in B's age.

Then, 3 x = the number of years in A's age.

Also, X — 8 = the number of years in B's age 8 years ago,

and 3 X — 8 = the number of years in A's age 8 years ago.

But A's age 8 years ago was 7 times B's age 8 years ago.
Whence, 3x-8 = 7(x-8).

Expanding, 3x — 8 = 7x — 56.

Transposing, — 4 x = — 48.

Dividing by — 4, x = 12, the number of years in B's age.

Whence, 3 x = 36, the number of years in A's age.

SIMPLE EQUATIONS. 55

Note 2. In Ex. 3, we do not say, "let v represenl B's ap'e," but
"let X represent the number of years in B's age."

4. A sum of money amounting to $4.32 consists of 108
coins, all dimes and cents j how many are there of each
kind?

Let X = the number of dimes. "■

Then, 108 — x= the number of cents-

Also, the X dimes-axe worth 10 x cents.
But the entire sum amounts to 432 cents.
Whence, lOx + 108 - x = 432.

Transposing, 9x = 324.

Whence, x = 36, the number of dimes ;

and 108 — x = 72, the number of cents.

PROBLEMS.

5. Divide 19 into two parts such that 7 times the less
shall exceed 6 times the greater by 3.

6. What two numbers are those whose sum is 246, and
whose difference is 72?

7. Divide 38 into two parts such that twice the greater
shall be less by 22 than 5 times the less.

8. Divide $22 among A, B, and C, so that A may
receive $2.25 more than B, and $1.75 less than C.

9. A is 5 times as old as B, and in 13 years he will be
only 3 times as old as B. What are their ages ?

10. B is twice as old as A, and 35 years ago he was
7 times as old as A. What are their ages?

11. A had one-third as much money as B; but after B
had given him $24, he had three times as much money
as B. How much had each at first?

12. A sum of money, amounting to $ 2.20, consists en-
tirely of five-cent pieces and twenty-five-cent pieces, there
being in all 16 coins. How many are there of each kind ?

56 ALGEBRA.

13. A is 68 years of age, and B is 11. In how many
years will A be 4 times as old as B?

14. A is 25 years of age, and B is 65. How many years
is it since B was 6 times as old as A ?

15. A man has two kinds of money ; dimes and fifty-cent
pieces. If he is offered $4.10 for 17 coins, how many of
each kind must he give?

16. Divide 76 into two parts such that if the greater be
taken from 61, and the less from 43, the remainders shall
be equal.

17. What two numbers are those whose sum is 13, and
the diiference of whose squares is 65 ?

18. Find two numbers whose difference is 6, and the
difference of whose squares is 120.

19. A is 14 years younger than B; and he is as much
below 60 as B is above 40. Required their ages.

20. A drover sold a certain number of oxen at $ 60 each,
and 3 times as many cows at $35, realizing $1485 from
the sale. How many of each did he sell ?

21. A man has $4.35 in dollars, dimes, and cents. He
has one-fourth as many dollars as dimes, and 5 times as
many cents as dollars. How many has he of each kind ?

22. A garrison of 4375 men contains 4 times as many
cavalry as artillery, and 7^ times as many infantry as
cavalry. How many are there '^f each kind ?

23. At an election where 5760 votes were cast for three
candidates, A, B, and C, B received 5 times as many votes
as A, and C received twice as many votes as A and B
together. How many votes did each receive ?

^ 24. Divide $ 115 among A, B, C, and D, so that A and
B together may have $ 43, A and C $ 65, and A and D $ 57

SIMPLE EQUATIONS. 67

25. A man divided $ 1G56 among his wife, three daugh-
ters, and two sons. The wife received 4 times as much
as either of the daughters, and each son one-third as
much as each daughter. How much did each receive ?

26. Divide $ 125 among A, B, C, and D, so that A and
B together may have $ 65, B and C $ 52, and B and D 1 54.

27. A man has 4 shillings m three-penny pieces and
farthings ; and he has 23 more farthings than three-penny
pieces. How many has he of each kind ?

28. Divide 71 into two parts such that one shall be 4
times as much below 55 as the other exceeds 37.

29. A square court has the same area as a rectangular
court, whose length is 9 yards greater, and width 6 yards
less, than the side of the square. Find the area of the
court.

30. Two men, 84 miles apart, setting out at the same time,
travel towards each other at the rates of 3 and 4 miles an
hour, respectively. After how many hours will they meet ?

31. Eind three consecutive numbers whose sum is 108.

32. In 7 years, A will be 3 times as old as B, and 8
years ago he was 6 times as old. What are their ages ?

(Let X represent the number of years in B's age 8 years ago.)

33. A sum of money, amounting to $ 24.90, consists en-
tirely of $ 2 bills, fifty -cent pieces, and dimes ; there are 5
more fifty-cent pieces than $2 bills, and 3 times as many
dimes as f 2 bills. How many are there of each kind ?

34. Find two consecutive numbers such that the difference
of their squares, plus 5 times the greater number, exceeds
4 times the less number by 27. •

35. Find four consecutive numbers such that the product
of the first and third shall be less than the product of the
second and fourth b} 9.

68 ALGEBRA.

36. A laborer agreed to serve for 32 days on condition
that for every day he worked he should receive $ 1.75, and
for every day he was absent he should forfeit $ 1. At the
end of the time he received $ 28.50. How many days did
he work, and how many days was he absent ?

37. A merchant has grain worth 5 shillings a bushel,
and other grain worth 9 shillings a bushel. In what pro-
portion must he mix 24 bushels, so that the mixture may
be worth 8 shillings a bushel ?

38. A general, arranging his men in a square, finds that
he has 43 men left over. But on attempting to add 1 man
to each side of the square, he finds that he requires 108
men to fill up the square. Kequired the number of men on
a side at first, and the whole number of men.

39. In a school of 535 pupils, there are 40 more pupils
in the second class than in the first, and one-half as many
in the first as in the third. The number in the fourth class
is less by 30 than 3 times the number in the first class.
How many are there in each class ?

40. A man gave to a crowd of beggars 15 cents each, and
found that he had 80 centf left. If he had attempted to
give them 20 cents each, he vvould have had too little money
by 10 cents. How many beggars were there ?

41. A tank containing 120 gallons can be filled by two
pipes, A and B, in 12 and 15 minutes, respectively. The
pipe A was opened for a certain number of minutes ; it was
then closed, and the pipe B opened; and in this way the
tank was filled in 13 minutes. How many minutes was
each pipe open ?

42. A grocer has tea worth 70 cents a pound, and other
tea worth 40 cents a pound. In what proportion must he
mix 50 pounds, so that the mixture may be worth 49 cents
a pound ?

IMPORTANT RULES. 59

VIII. IMPORTANT RULES IN MULTIPLICA-
TION AND DIVISION.

78. Let it be required to square a + b.

a + b
a + b

a^ 4- db

ab + b^

Whence, (a + by = a'^ + 2ab + b\

That is, the square of the sum of two quantities is equal to
the square of the first, plus twice the product of the two, plus
the square of the second.

Example. Square 3 a + 2 be.

We have, (3 a + 2 6c)2 = (3 a)2 + 2 x 3 a x 2 6c + (2 bey
= 9 a2 + 12 a6c + 4 dV, Ans.

79. Let it be required to square a — b.

a -

-b

a -

-b

a'-

- ab

-ab + W

Whence, (a -by = a'-2ab + b\

That is, the square of the difference of two quantities is equal
to the square of the first, minus twice the product of the two,
plus the square of the second.

Example. Square 4 cc — 5.

We have, (4 a; - 5)2 =(4a;)2 - 2 x 4x x 5 + 52

= 16 a;2 - 40 X + 25, Ans.

60 ALGEBRA.

80. Let it be required to multiply a + 6 by a — 6.

a + 6
a — b

^

a^ + ab
-ab-b^

Whence, (a + b)(a — b) = a^ — b^.

That is, the prodtict of the stem and difference of two quanti-
ties is equal to the difference of their squares.

Example. Multiply 6 a^ + 6 by 6 a^ — &.

We have, (6 a^ + b) (6 a^-b) = (6 a^y - b"^ = 36 a* - b\ Ans.

81. In connection with the examples of the present
chapter, a rule for raising a monomial to any power whose
exponent is a positive integer will be found convenient.

Let it be required to raise 5 a%^c to the third power.

We have, (5 a^b^cY = 5 a-b^c x 5 a^b'^c x 5 a^b^c = 125 a*6V.

We then have the following rule :

liaise the numerical coefficient to the required power, and
multiply the exponent of each letter by the exponent of the
required power.

EXAMPLES.

82. Find by inspection the values of the following :

1. {x + ^f. 9. (8 + 3m3n2)2.

2. (a -3)1 10. (ab'-\-2a^¥f.

3. (6a -5 6)2. 11. {&xy-lxzy.

4. {2xy + o,y_ 12. {^a^' + Wbcy.

5. (3m + 4n)(3m-4n). 13. {^ xxf + 2 z^i^ xy"" - 2 z^).

6. {1-2 ay. 14. {lab-bcdf.

7. (5a^ + 8)(5a^-8). 15. (Q>x' + llf)(<o:(? -llf)

8. (a*-%ay. 16. {^a' + ^aj.

IMPORTANT RULES. 61

17. (7m3+12n)(7?/i3-12?i). 20. (3a^ + 4b^y.

18. {Sx' + Ty^y. 21. (Sxp-Sx^y.

19. (13 a*x- 6 byy. 22. (a-' + a') (a^ ~ a'y
\23. Multiply a + b + chy a i- b — c.

(a+ b + c)(a-^ b-c) = [(a + 6)+c][(a + &)-c]

= (a + 6)2 - c2 (§ 80)

= a2 + 2 a6 + 62 _ c2, ^„s. (§ 78)

24. Multiply a+ b — chy a — b -\- c.

(a + 6 -c)(a - 6 + c) = [a +(6 -c)][a -(6-c)]

= «'-(?>-c)2 (§80)

= a'- - (62 - 2 6c + c2) (§ 79)

= a2 — 62 + 2 6c — c2, A71S.
Expand the following :

25. (a + b + c){a-b + c). 28. (a^ + a - l)(a2- a + 1).

26. (x-y-{-z)(x-y-z). 29. (a;- + a; - 2) (a;2 _ ^ _ 2).

27. (a + 6 + c)(a -b-c). 30. (1 + a + &) (1 - a - b).

31. (ar^ + 2 a; + 1) (a;- - 2 a; + 1).

32. (a + 2b-3c){a-2b + 3c).

33. (a^ + ab + b^ {a" - ab + b^).

34. (3a; + 4?/ + 2e)(3a;-42/-22).

83. We find by multiplication :

a; + 5 a; — 5

x+3 x-3

a;2 + 5a;
+ 3x

+ 15

x' + d,x

X + b
x-3

+ 15

x^ + ox
-3x

-15

7?

— 5x

-3x

+ 15

X?

-Sx

+ 15

X

-5

X

+ 3

^

— 5 a;

+ 3a;

-15

a;2 + 2x-15 a;2-2a;-15

62 ALGEBRA.

In these results it will be observed that :

I. The coefficient of x is the algebraic sum of the second
terms of the multiplicand and multiplier.

II. The last term is the product of the second terms of
the multiplicand and multiplier.

. By aid of the above laws, the product of any two binomials
of the form x + a, a; + 6 may be written by inspection.

1. Required the value of {x + 8) (x — 5).
The coefficient of x is + 8 — 5, or 3.

The last term is 8 x (— 5), or — 40.

Whence, (x + 8) (x - 5) = x2 + 3 a; - 40, Ans.

2. Required the value of (a — 6 — 3) (a — & — 4).
The coefficient of a — 6 is — 3 — 4, or — 7.

The last term is (—3) x'(— 4), or 12.

Whence, (a-b- 3)(a - 6 - 4) = (a - 6)2 - 7 (a - 6)+ 12, Ans.

EXAMPLES.
Find by inspection the values of the following :

3. (a; + 6) (a; + 4). 14. (a + b -7)(a + b + 8).

4. (x-2)(x + 3). 15. (x-5a)(x-lla).

5. (a; - 10) (a; - 1). 16. (x + y)(x -2y). .

6. (a; + 5) (a: - 6). 17. (a + lib) (a - 6b).

7. (a + l)(a + 9). 18. {a + 7 x)(a + 5x).

8. (a -7) (a +4). 19. (x -y - A)(x-y + 10).

9. (m + 5)i7n-l). 20. (x-llz)(x + 9z).

10. (ar* - 7) (a;2 - 2). 21. (x' -\-3y)(x' + 8y).

11. {n^ + S)(n^-10). 22. (x^ - 9 m') (x" - 6 tnr).

12. (ab + 2) (ab + 11). 23. (ab + 8 cd) (ab - 12 ccl).

13. (a^-12)(a^-3). 24. {x + y + 12){x + y -9).

IMPORTANT RULES. 63

84. We have by § 80,

_ ^2 a^ — b"^

=1 a — b, and = a + 6.

a + b a-b n/\u''

That is, if the difference of the squares of two quantities bey-^ 'HA fV^^
divided by the sum of the quantities, the quotient is the dif-
ference of the quantities.

If the difference of the squares of two quantities be divided
by the difference of the quantities, the quotient is the sum of
the quantities.

1. Divide 16 a^b* - 9 by 4 aft^ + 3.

We have, 16 a^-b* = (4 ab^y. (§ 81)

Whence, 16 a^b* - 9 ^ 4 ^^2 _ 3 ^ns.

4 fflft2 + 3

U^

EXAMPLES.

Write by inspection the values of the

1 following:

2.

x + 1

5.

25a^-36

5a2 + 6

g 1-64 m V

1+8 mn

3.

4-a2
2-a

6.

9x''-16y'
3x + 4y

4 a2&« - c"
2ab^-c'

4.

16m2-49
4 ??i — 7

7.

25 a^ - b*
ba-W

jQ 49m2-100n''
7 m - 10 ?|8

11

Sly^-196x*

1 q 144 x'y^ - 169 2«

9y+Ux'

^'^- 12

xy^ - 13 2^

12.

121 b^c^- 64. a^

11 225,

xio_64 6'2c8

llbc + 8a

^*- 15

a* + 8 6«c*

85. We find by actual division :

t±]l.^a'-ab-\-b\
a-\- b

o? — b^

and = a^ + a6 + b^.

a — b

^'

64 ALGEBRA.

That is, if the sum of the cubes of two quantities be divided
by the sum of the quantities, the quotient is the square of the
first quantity, minus the product of the two, plus the square of
the second.

If the difference of the cubes of two quantities be divided by
the differeyice of the qxiantities, the quotient is the square of the
first quantity, plus the product of the two, plus the square of
the second.

1. Divide l+Sa^ by 1-f- 2a.

We have, l+^^l+(2ay

= 1 - 2 a + (2 a)2
= 1 - 2 a + 4 a2, Ans.

2. Divide 27 ar* - 64 / by 3 x - 4 2/.

We have, 21x^-6iy^ ^ i3xy-(4yy

3x — 4y Sx — iy

= (3x)2 + (3x)(4 2/) + (4 2/)2
= 9x^ + \2xy + IQ y\ Ans.

EXAMPLES.
Find the values of the following :

3 ^'-1 8 ^'^' - ^' 13 8a:^-125y«
a — 1 a^b — c? 2x — 5y^

4 1±^. 9 1 + 64 m\ j^ a^b'' + 512 c^
"l+a; l + 4m ' ab + Sc '

g m^ + 8 jQ 216 -af' jg 64 m^n^ + 343

m + 2 6 — a; 4 mn + 7

6 ^^-<^' 11 g' + 125 jg 729 a''- 125 a^

3 — a a4-5 9a^ — 5a;

- x^ + y^ j2 l-343a'^6« ^^ 512a;Y + 272«

■ a^ + / ■ l-7a62 ■ ' ^^y-^^z" '

a

-\-b

a*

-b*

a

-b

a'

+ b'

a

+ b

a'

-b'

IMPORTANT RULES. 65

86. We find by actual division :

~ = a^ — arb + ab- — b^,

= 0? -\- orb + ab- + b^,

= a'^ — a^b + orb- — aW + b*,

= a^ + a^b + a-b- + aW + b^ ; etc.
a — b

In these results we observe the following laws :

I. The number of terms is the same as the exponent of
a in the dividend.

II. The exponent of a in the first term is less by 1 than
its exponent in the dividend, and decreases by 1 in each
succeeding term.

III. The exponent of b in the second term is 1, and
increases by 1 in each succeeding term.

IV. If the divisor is a — b, all the terms of the quotient
are positive ; if the divisor is a-\-b, the terms of the quo-
tient are alternately positive and negative.

87. The following principles are of great importance.
If n is any positive integer, it will be found that :

I. a" — 6" is always divisible by a — b.

Thus, a^ — b^, a^ — W, a"* — b*, etc., are divisible by a — 6.

II. a" — 6" is divisible by a + b if n is even.

Thus, a^ — 6^, a* — b*, a^ — b^, etc., are divisible by a+b.

III. a" + 6" is divisible by a -\-b ifn is odd.

Thus, a^ + b^, a^ + b^, a' + b\ etc., are divisible by a + 6.

IV. a" + 6" is divisible by neither a + b nor a — b if n
is even.

Thus, a- + b^, a* + b*, a^ + b% etc., are divisible by neither
a + b nor a — b.

66 ALGEBRA.

88. 1. Divide oH — V by a — h.
Applying the laws of § 86, we have,

a—h

2. Divide 16 x^ - 81 by 2 x + 3.

* We have, IMzlM = (2^)lzi3!
2x + 3 2x + 3

= (2 x)3 - (2 x)2 X 3 + (2 a;) X 32 - 33

= 8x3-12x2 + 18x - 27, ^ns.

EXAMPLES.

Find the values of the following :

3. r- 9. — 15. — — .

2a — h

4. ^-1. 10. Izil^. 16 ^^^''-^'.

Saj + y ■

a«-243a^

a

+ 1

a^

-1

a;

-1

a«

-&«

a

+ 6

1-

-a^

1-

— a;

a*

-6»

0?

-62

x''

y + 2^5

2-a;

10

l-16a^

l+2a

11.

aJ-hb'

a-\-b

12.

l-mJ

1 — m

13.

32 + a«

2 + a

Id

m* — ri®

c g" — o" J, g- -t- 0' iw

a + 6 a + &

6. ^^^i^. 12. ^-^'. 18.

a — 3a;

81a^-256&^
3a-46

^ gs-fts ^g 32 + a» jg 243a^ + 32y^

" ""' ■ ~ * * 3x-{-2y

2Q 128m^-7t"
x^y -\-z^ m + n 2 m — n^

FACTORING. 67

IX. FACTORING.

89. To Factor an algebraic expression is to find two or
more expressions which, when multiplied together, will
produce the given expression.

90. A Common Factor of two or more expressions is an
expression which will exactly divide each of them.

91. A monomial can always be factored ; thus,

12 a^hc^ = 2x2x3xaxaxax&xcxc.

It is not always possible to factor a polynomial; but
there are certain types which can always be factored, the
more important of which will be taken up in the present
chapter.

92. Case I. When the terms of the expression have a
common monomial factor.

1. Factor Uxy*-35x^y^

Each term contains the monomial factor 7 xy'^.

Dividing the expression by 7 xy'^, the quotient is 2y^ — 5 7?.

Whence, 14 xy* - 35x^2/2 = 7 xy\2 1 - 6 x2), Ans.

EXAMPLES.
Factor the following :

2. a3 + 4a. ' 7. 12a*-20a3 + 4a2.

3. 6iB*-14a;3 g^ a*6V + a^ftV + a'&c^-

4. 30m2-5m''. 9. 12 a^/ + 24 a;?/^ - 42 a;y.

5. \ha%''^^ah\ 10. 14 a^6* + 21 a^ft^ _ 49 ^352

6. 56a^2/='-32a;V. 11. 81 m^n + 54 mV + 9 m'n^.

12. 48xV-144a^^ + 108ar'^*.

13. 70 aV - 126 aH'' - 112 a^^.

68 ALGEBRA.

93. Case II. When the expression is the sum of two bino-
mials lohich have a common binomial factor. .

1. Factor ac — bc + ad — bd.

By § 92, {ac - be) + (ad - bd) = c(a -b)+ d(a - b).
The two binomials have the common factor a — b.
Dividing the expression by a — 6, the quotient is c + (?.
Whence, ac — be -\- ad — bd = (a — b) (c + d), Ans.

If the third term of the given expression is negative, as
in the following example, it is convenient to enclose the
last two terms in a parenthesis preceded by a — sign.

2. Factor 6 a^ - 15 ar - 8 x + 20.

6 a;3 - 15 x2 - 8 X + 20 = (6 a;3 - 15 x2) - (8 a; - 20)
= 3x2(2x-5)-4(2x-5)
= (2x-5)(3x2-4), Ans.

EXAMPLES.

Factor the following :

3. ab + an -f- 6m + mn. 9. 3ci^-\-6x^ + x-\-2.

4. ax — ay + bx — by. 10. 10mx—15nx—2m-{-3n.

>

5. ac — ad — be -\- bd. 11. a^x + abcx — a^by — b^cy.

6. a^ + a^ + a -{- 1. 12. a^bc — ac^d + ab^d — bcd^.

7. 4:a^-5ay'-4:X-\-5. 13. SOa^- 12a2 - 55a + 22.

8. 2 + 3a-Sa'-12a^ 14. 56 - 32 a; + 21 ar^ - 12 a;^

15. a^b^ + a'bcd^ + ab-c''d + <^d\

16. 3ax — ay — ^bx + 3by.

17. 4 a^ + a;^ — 16 a;?/ — 4 2/^.

18. 20 ac + 15 6c + 4 ad + 3 6d

19. 16 mx — 56 my + 10 nx — 35 ny.

20. 45 a3 - 20 a^ft^ _ 63 a6 + 28 b\

FACTORING. 69

94. If an expression can be resolved into two equal
factors, it is said to be a perfect square, and one of the
equal factors is called its square root.

Thus, since 9 a'^lr is equal to 3 a^6 x 3 a^h, it is a perfect
square, and 3 a'h is its square root.

Note. 9 a*h'^ is also equal to ( — 3 a%) x ( — 3 a%) ; so that — 3 a%
is also its square root. In the examples of the present chapter, we
shall consider the positive square root only.

95. The following rule for extracting the square root of
a perfect monomial square is evident from § 94 :

Extract the square root of the numerical coefficient, and
divide the exponent of each letter by 2.
Thus, the square root of 25 a*b^c- is 5 a'^b^c.

96. It follows from §§78 and 79 that a trinomial is
a perfect square when its first and last terms are perfect
squares and positive, and the second term twice the product
of their square roots.

Thus, in the expression 4 .'k^ — 12 xy + 9 y^, the square
root of the first term is 2 x, and of the last term 3 y ; and
the second term is equal to 2 (2 x) (3 y).

Whence, Aa^ — 12xy + 9y^ is a perfect square.

97. To find the square root of a perfect trinomial square,
we simply reverse the rules of §§ 78 and 79 :

Extract the square roots of the first and last terms, and
connect the results by the sign of the second term.

Thus, the square root oi 4: x^ — 12 xy -\- 9 y"^ is 2x — d y.

98. Case III. When the expression is a perfect trinomial
square (§ 96).

1. Factor a^ + 2 ab^ + b*.

By § 97, the square root of the expression is a + b"^.

Whence, a^ + 2 ab"^ + b* =(a + b'^y =ia + b'^) (a + b'^), Ans.

70 ALGEBRA.

2. Factor 25 x" -4S)xf-\- 16 y\

The square root of the expression is 5 x — 4 y^.

Whence, 25 x^ - 40 xy^ + \Qy^ z=(bx - iy^)^

= (5x — 4?/3)(5x — 4?/8), Ans.

Note. The expression may be written 16 y^ — 40 xy^ + 25 x^ ; in
which case, according to the rule, its square root is Ay^ — bx.
Thus, another form of the result is

162/6 - 40XJ/3 + 25x2 =(4 ?/3 _ 5x)(4j/' - 5a;).

EXAMPLES.

Factor the following :

3. 'm? + 2mn-\-n\ 15. Ua%^ -\-lQdbcd-{- (?d\

4. a'-2ab + h\ -^ 16. 100 a^ - 60 a;« + 9 aj*.

5. 9 + 6a; + a^. 17. 49 m* + 112 mV + 64 w«.

6. a^-Sa + lB. 18. 121 a^ft" + 132 a&c^ + 36 c".

7. 49 ic^ 4- 14 «?/ + 2/2. 19. 144«y-120a;y+25a;y.

8. m?-l(imn-\-25n\ 20. Q>4.a?lJ'+VJQahh+121h''c\

9. 4 a* - 4 a^ft^c + & V. 21. 49a^2/''- l^Sa;^^^ + 144z^

10. m^o^ + 18 mx + 81. 22. 36aV-156a^a;3+169a-V.

11. 4 a* - 20 aa; + 25 a^. 23. (a + &)2-4(a + 6) + 4.

12. 9 a2 + 42 a6 + 49 ft''. 24. (a; - y)^ + 10 (a; - y) + 25.

13. 81ar'-72a^ + 162/2. 25. 16(a + a;)2 + 8(a4-a;) + l.

14. a^ + 12a;V + 362/V. 26. 4(a - 6)2-12(a- &) + 9.

99. Case IV. When the expression is the difference of two
perfect squares.

By § 80, a^ - 6^ = (a + b) (a - b).

Hence, to obtain the factors, we reverse the rule of § 80 :

Extract the square root of the first sqxiare, and of the second
square; add the results for one factor, and subtract the second
result from the first for the other.

FACTORING. 71

1. Factor 36 a- - 49 b\

The square root of 36 a- is 6 a, and of 49 b* is 7 6^.

Whence, 36 a^ - 49 6^ = (6 a + 7 62) (6 a - 7 62) , Ans.

EXAMPLES.
Factor the following :

2. a--b\ 8. 49m*-16?r. 14. 144 m V - 49.

3. x2_i. 9. 25 a^- 64 6V. 15. 36a«-169a^.

4. 9-m-. 10. lOOarV-9 2^ 16. 81a;'«-196i/V.

5. 16ar-?/l 11. 64?R*-81n«. 17. 64 a^i* - 225 ci".

6. 4.0? -25. 12. 121a-62-4c2dl 18. 169-1440^?/".

7. l-36a26-. 13. 81a3«-100/. 19. 196 a^x^ - 121 6y.

20. Factor (2 a; - 3 yf - (x - yf.
We have, (2x—3yy-(x-yy

= [(2x-32/) + (x-2/)][(2x-3y)-(x-2/)]
= (2x-32/4-x-?/)(2x-32/-x + 2/)

= (3x — 4:y)(x — 2y), Ans.

)■'•■

Factor the following :

21. (a + 6)2-cl 28. (a + 6)^ - (c - d)^.

22. (m-ny-x'. j^ 29. (a - a;)^ - (6 - ^)2.

23. a'-ib-cy. ^ "30. (a; + 2/)' - (m-f- n)2.

24. ar'-(2/ + 2)2. 31. (8 a - S)'' - (3 a + 7)^.

25. m?-{n-py. 32. (4 a; + l)^ - (a; + 6)2.

.26. {lx-2yy-y\ 33. (7a - 56)2 - (5a- 26)1

27. (a - 6)- - (a; + 2/)'. 34. (9 a; + 8 yy - (2 x - 8 yy.

A polynomial may sometimes be expressed in the form
of the difference of two perfect squares, when it may be
factored by the rule of Case IV.

^

72 ALGEBRA.

35. Factor 2 mn -}-m^ — 1 -\- n^.

Since 2 mn is the middle term of a perfect trinomial square whose
first and third terms are m- and n'^ (§ 96), we arrange the given ex-
pression so that the first, second, and last terms shall be grouped
together, in the order m^ + 2 mn + n^ ; thus,

2 mn + m^ - 1 + n^ = (m^ + 2 mn + n^) — 1

= (m + n)2 - 1, by Case III.
= (m + n + 1) (m + n — 1), Ans.

36. Factor 12 ?/ + ar' - 9 ^/^ - 4.

We have, 12 y + x^ - 9 2/2 - 4 = x2 - 9 ?/2 + 12 j/ - 4

= a;2 _ (9 y2 _ 12 y + 4)
= x2 - (3 y - 2)2, by Case III.
= [x + (3y-2)][a;-(3y-2)]
= (x-l-3y-2)(x-3j/ + 2), ulns.

37. Factor a'- c' + b^-d' -2cd-2 ah.

We have, a2 - c2 + &2 _ ^ _ 2 cd - 2 a6

= a2 - 2 a6 + ft2 _ c2 - 2 cd - (Z2
= (a2 _ 2 a6 + 62) _ (c2 + 2 cd + d^)
= (a - 6)2 - (c + d)2, by Case III.
= [(a_6) + (c + d)][(a-6)-(c + d)]
= (a — 6 + c-|-d)(a — 6 — c — d), .4ns,

Factor the following :

38. a2-2a6 + 62-c2. 43. 2 77171 - n^ + 1 - ml

39. w? + 2mn + n--p\ 44. Oa^ -24 a& + 1662-4 c^.

40. a'-x'-2xy-f?) 45. 16ar^-42/'^-f 203/2-25zl

41. a^ — t/'^ — 2^ + 2 2/2. 46. 4 n^ + m^ — a^ — 4 mn.

42. 6''-4-f 2a6 + al 47. 4a=^-66-9-&l

48. 10 a:?/ -9 2- + 7/2 + 25 ar'.

49. a^ - 2 a6 + &- - 0-2 + 2 cd - dl

50. a^-b- + x^-f- + 2ax + 2 by.

FACTORING. 73

51. x^ + m^ — 'if — n^ — 2 mx — 2 n?/.

52. 2 a;^ - a^ -|. ^2 _ 2 a6 - 6^ + ^/S.

53. 4 a^ + 4 a6 + 6^ - 9 c^ + 12 c - 4.

54. 16y^-36-8xy-z^ + x^-12z.

55. m^ - 9 n^ + 25 a^ - 6^ - 10 a?/i + 6 5/i.

100. Case V. When the expression is a trinomial of the
form x"^ + ax-{- b.
We have by § 83,

(x + 5)(x + 3) = x' + 8x + 15,

{x-5){x-3) = x^-8x + 15,

(x-{-5)(x-3) = a^ + 2x-15,

and {x-5)(x + 3) = x'-2x- 15.

In certain cases it is possible to reverse the process, and
resolve a trinomial of the form x^ + ax -j-b into two binomial
factors.

The first term of each factor will obviously be x ; and to
obtain the second terms, we simply reverse the rule of § 83.

Find two numbers whose algebraic sum is the coefficient of x,
and whose product is the last term.

1. Factor x^ -^Ux + 4:5.

We find two numbers whose sum is 14, and product 45.
' By inspection, we determine that the numbers are 9 and 5.
Whence, x^ + 14 x + 45 = (a: + 9) (x + 5), Ans.

2. Factor a^ — 5 x + 4.

We find two numbers whose sum is — 5, and product 4.
Since the sum is negative, and the product positive, the numbers
must both be negative.

By inspection, we determine that the numbers are — 4 and — 1.
Whence, x^ — 5 x + 4 = (x — 4) (x — 1), An/i.

74 ALGEBRA.

3. Factor ic^ + 6 X - 16.

We find two numbers whose sum is 6, and product — 16.

Since the sum is positive, and product negative, the numbers must
be of opposite sign, and the positive number must have the greater
absolute value.

By inspection, we determine that the numbers are + 8 and — 2.

Whence, x2 + 6 x - 16 = (x + 8) (x - 2), Ans.

4. Factor a^ — a; — 42,

We find two numbers whose sum is — 1, and product — 42.
The numbers must be of opposite sign, and the negative number
must have the greater absolute value.

By inspection, we determine that the numbers are — 7 and + 6.
Whence, x^ — x — 42 = (x — 7) (x + 6), Ans.

Note. In case the numbers are large, we may proceed as follows :

Required the numbers whose sum is — 26, and product — 192.

One number must be +, and the other ~.

Taking in order, beginning with the factors + 1 x — 192, all possible
pairs of factors of — 192, one of which is + and the other — , we
have :

+ 1 X ~ 192,

+ 2 X - 96,

+ 3 X - 64,

+ 4 X - 48,

+ 6 X - 32.

Since the sum of -f 6 and — 32 is — 26, they are the numbers
required.

EXAMPLES.

Fac

itor the following :

5.

a^ + 6 » + 8.

11.

x'-x-G.

6.

a)2_13a; f 22. '

12.

a^ + 10 a; + 9.

7.

c^^Qx-7.

13.

a- - 7 a - 44.

8.

ar^-4cc-21.

14.

a- + a - 2.

9.

0^2 -lite + 24. —

15.

m^ + 11 m + 30.

10.

CC2 + 8 CB - 20.

16.

n2-7w + 6.

4

' FACTORING. 75

17. x^ + Zx- 40. "'^ • 31. 2^ - 21 2 + 110.

18. ?/2 + 18y + 77. 32. x^ + 17r^-84.

19. a^ - 15 a + 54. 33. a* + 25 a^ + 150.

20. m2-2?;i-48. 34. m^-bw?-3Q.

21. c2 + 15c + 36. 35. w^ + lOn^-QG.

22. 0^-12 a; + 32. 36. a^V - 19 a;?/ + 84.

23. a^-6a;-55. 37. a^i^ .4. 28 a& + 160.

24. ?r + 2w-63. 38. a;V - 27 a^y + 50.

25. m^- 18 m + 72. 39. aV + 5aV-126.

26. a2-3a-70. 40. m=^w« - 11 mn^ - 152.

27. a^ + 4a;-96. 41. (a + &)2+ 23(a + &)+ 60.

28. x2 + 24a; + 95. 42. {x -yf-\-Z(x-y)-1^0.

29. &2_io5_24. 43. (a -6)2- 22 (a -6) +112.

30. c2 + 20c + 84. 44. (x + yy-2{x + y)-UZ.

45. Factor a;^ + 5 a&a; - 27 a%-.

We find two quantities whose sum is 6 ah, and product — 27 a^ftz.
By inspection, we determine that the quantities are — 3 ah and 9 ah.
Whence, x^ + 6 a6x - 27 a^ft'^ =:(x-3a6)(x + 9a6), ^ns.

46. Factor 1 - 3 a - 88 a^.

We find two quantities whose sum is —3 a, and product — 88 a^.
By inspection, we determine that the quantities are 8 a and —Wa.
Whence, 1 - 3 a - 88 a2 := (i + 8 a) (1 - 11 a), Ans.

Factor the following :

47. a^ + 12 a6 + 35 &2. 51. a? + bam-mm'.

48. a;2 ~ 11 aa; + 28 al 52. m' + 16 m?i + 48 w''.

49. a;2 + 4xy-5/. 53. a;^ - ma; - 12 m*.

50. l-2a-3a'=. 54. 1- 14a + 33 a''.

76 ALGEBRA.

55. a^ - 4 a6 - 60 h\ 61. 1 + 18 ah + 80 a'h-.

56. l + .r-72a^. 62. .^^^ + 7 .r// - 60 t/^.

57. a? - 15 ici/ + 50 y\ 63. a'b' + 16 ahc + 28 c\

58. ic2 + 20 a-T + 99 al 64. ar - 21 a;yz + 108 yV.

59. m^ — 16 nin + 15 w'. 65. 1 + 11 a*?/ — 26 .t-?/^.

60. a-~ab-20h\ 66. a" - 6 a^fec^ - 160 6V.

'cy'lOl. If an expression can be resolved into three equal
factors, it is said to be a perfect cube, and one of the equal
factors is called its cube root.

Thus, since 21 a^h^ is equal to Za-h x Sa^b x 3a-b, it^is a
perfect cube, and 3 arb is its cube root.

102. The following rule for extracting the cube root of a
perfect monomial cube is evident from § 101 :

Extract the cube root of the numerical coefficient, and divide
the exponent of each letter by 3.

Thus, the cube root of 125 a^ftV is 5 a-b^c.

103. Case VI. When the expression is the sum or differ-
ence of two perfect cubes.

By § 85, the sum or difference of two perfect cubes is
divisible by the sum or difference, respectively, of their
cube roots.

In either case, the quotient may be obtained by aid of the
rules of § 85.

1. Factor a^ + l.

The cube root of a^ is a, and of 1 is 1 ; hence, one factor is a + 1.
Dividing a^ + 1 by a + 1, the quotient is a- — a + 1 (§ 85).
When.e, a^ + 1 ={a + l)(a^ - a + 1), Ans.

2. Factor 27 ar' - 64 f.

The cube root of 27 x^ is 3 x, and of 64 yMs 4 y (§ 102).
Hence, one factor is 3 a; — 4 ?/.

Dividing 27 x^ — 64 y^ by 3 x — 4 y, the quotient is 9 x^ + 12 xy + 16 ?/*
(§ 85).

Whence, 27 x^ - 64 y^ = (3 x - 4 y) (9 x- + 12 xy + 16 y^) , Ans.

^

FACTORING. 77

3.

iiv^ + n^.

9.

64 x^ + 1.

4.

a^ - b\

10.

l-126al

5.

a'-l.

11.

27a;3-8y^.

6.

aj3 _ y3^3^

12.

8 a'b' + 125.

7.

a^ + x".

13.

216 - m^

8.

1 + 7ft«.

14.

125-6Axhj

EXAMPLES.

Factor the following:

15. m*^ + 3437i3.

16. 125 &== - 216 c^.

17. 34:3 m' + S a^.

18. 27a«+343 6«.

19. 512 ar* + 27 2/V.

20. 64tt36«-729cl

104. Case VII. TFAen the expression is the sum or differ-
ence of two equal odd 'powers of two quantities.

By § 87, the sum or difference of two equal odd powers
of two quantities is divisible by the sum or difference,
respectively, of the quantities.

In either case, the quotient may be obtained by aid of the
rules of § 86.

1. Factor a' + h^.

By § 87, one factor is a + 6.

Dividing a^ + 6^ by a + 5, the quotient is

a* - a^b + a-^62 _ ^^3 + ^4. (§ 86)

Hence, a^ + h^={a + h) (a* - a% + a^b'^ - ab^ + b*), Ans.

EXAMPLES.
FEictor the following :

2. ct^-f. 6. a' + bl 10. l+32ar'.

3. a' + l. 7. l-x'. 11. 243m^-l.

4. l-m^ 8. m' + l. 12. a;^ - 128.

5. x^f+z". 9. 32 -a^ 13. 32a^-|-243 6^

105. By application of the rules already given, an ex-
pression may often be resolved into more than two factors.

78 ALGEBRA.

If the terms of the expression have a common monomiai
factor, the method of Case I should always be applied first.

1. Factor 2 axhf — 8 axy^.

We have, 2 ax^if- — 8 axy^ = 2 axy''-{x'^ — 4 j/^), by Case I.
"Whence by Case IV,

2 ax'V^ — 8 axi/ = 2 ax]p'{y. + 2 y) (x — 2 y) , Ans.

If the given expression is in the form of the difference
of two perfect squares, it is always better to apply first the
method of Case IV.

2. Factor a^ — b^

We have, «« - fe^ = (a3 + feS) (a^ - b"^), by Case IV.
Whence by Case VI,
a6 _ 56 ^ (a + b) (a2 -ab+ b"^) (a - b) (a^ + ab + 6^) , Ans.

3. Factor sc^ — ?/*.

We have, x^ - y^ = (x* + ?/*) (x* - y*), by Case IV

= (x* + 2/*) (x2 + 2/2) (x + ?/) (a; - y) , Ans.

* MISCELLANEOUS AND REVIEW EXAMPLES.
106. Factor the following :

1. 35 a'b^ + 98 a'^b^ - A9 a'b. 10. 4.a'b^ + 4:a%^

2. 25 ahn*- SI bhA 11. a^ _(. 15 «& + 56 &".

3. ar^ + 11 a; + 18. 12. x'y^ - 23 xij + 132.

4. a^hc+cucH-ab-d-bcd?. 13. 108 x* - 36 .r^ + 3 .r'.

5. 6a^-6a;^. 14. Q4.a?b -121 u'W.

6. 49 m^ + 56 mji + 16 rr. 15. a;« - 1.

7. a^- 10 a + 24. 16. a.-^ + x-y + a-/ + 2/3_

8. ar^ + 17 x^ - 38 a;. 17. aVj' -3ab'' -ISO.

9. a2_(^, + c)2. 18. 2a^ + 20 a;?/ - 78 /.

FACTORING. 79

19. 30 a:^-55 x'-\-65 x-«-20 x\ 25. 27 a' - 64 a^.

20. 1 - a«. 26. 32 x^ + 2/'"-

21. 16 x* - 1. 27. 8 a'^b - 72 a-i^ + 162 ab^

22. 64 a-62 _ 80 abc + 25 c'. 28. 1-11 mn - 60 m^nl

23. 15ac+18ad-356c-42 M. 29. (a; - ?/)2-(m - n)2.

24. 100 x« - 49 y'z'. 30. (1 + ?i2)2_ 4 ,^2.

31. 64 a^?/3^- 56 a;Y2* + 72x^/2*.

32. 3 a'b^ - 3 a&^ 50. 9 or + 25 / - 16 z' + 30 a;y.

33. m* - 81. 51. 343 m^ + 216 w^.

34. 8 x^/ + 125. 52. (9 a^ + 4)^ - 144 al

35. (m+n)2+7(m + w)-144. 53. (T^ + x-df-d.

S6. a'x"- 15 abxy- 54: by. 54. (a2-2a)-+2(a2-2a) + l.

37. 25 x' + 110 xy + 121 ^/l 55. a'b^ + aV - i^x^ - ory.

38. 4 a^ - 8 a^ - 2 a^ + 4 al 56. of - 256.

39. (5a;-8?/)2-(4x-92/)2. 57. 36a2~4&2_49c2 + 28&c.

40. 5a;9 + 5r=. 58. m*-625.

41. (a- + 9)2 - 36 al 59. (a.-2+3 a;)^ +4 (.t^+S a;) +4.

42. a;^ - (a; + 2)'. 60. a« - 7 a^ - 8.

43. aV-4&262-9a2d2_,_36j2^2 q^ 27 a« - 1000 SV^.

44. (a;2-5a;)2-2(a;2_5^^)_24. 62. 128 - ?7i^.

45. 16 x* - 72 ctfy' + 81 1/^ 63. 2 a^&c - 2 b'^c - 4 6^-2 &c^

46. a''-2a3 + l. Qi, (a'+7ay+4:(a'+7a)-96.

47. 64 - x^. 65. x^" + 2 a;^ -I- 1.

48. 45 a^+18 x*+60a^+24 a;^. 66. (a^ - 4)^ - (a; + 2y.

49. 9 (m - w)2 - 12 (7n - n) + 4. r^7. (a'-b- + (ff - 4 a'c".

68. "Resolve a;^ — y* into two factors, one of which is x-^y.

69. Eesolve a^ — b'^ into two factors, one of which is a — b.

80 ALGEBRA.

70. Resolve x^ + y^ into two factors by the method of § 104.

71. Resolve x' + y^ into three factors by the method of

§ 103.

72. Resolve 1 — m^ into two factors by the method of § 104.

73. Resolve a^ — 1 into three factors by the method of

§ 103.

74. Factor 3 (m + n)- - 2 (m^ - rr).

3(m + n)'^ — 2(m2 — n^) = 3(m + n)^ — 2(m + n) (m — n)
=:(m + n)[3(m + ?i)— 2(m — »»)]
= (m + ?i) (3 m + 3 « - 2 jw + 2 n)
= (m + n) (m + 5 re) , Ans.

75. Factor (a + bf - (a - 6)^
By the method of § 103, we have

(a + 6)3- (a -6)3

= [(a + ft) - (a - 6)] [(a + 6)^ +(a + b)(a - b)+(a - 6)2]
= (a + 6 - ffl + 6) (a2 + 2 ffl6 + 62 + a2 - 62 + a2 - 2 a6 + 62 )
= 2 6(3a2 + 62), ^?is.

Factor the following :

76. (m -xf + Sx^. 84. a--62+a;''-2/2+2aa;+26?/.

77. a^ — (a — by. 85. (a; — my — x{x^ — m^).

78. 5 (a^ - 2/-) + 4 (.r - y)^ 86. (x + ?/)3 - (a: - t/)^.

79. (a^ + &^) - 2 a6 (a + b). 87. a^" - 1.

80. a2+6--c2-d2 + 2a6-2cd. 88. .t^^ + a;* - a^ - 1.

81. (x + ly + (.^ - 1)^ 89. (a' - 1) - (a - 1)^.

82. (x" + /) + .1- (a; + yy. 90. (3 m- 2)^ + (2 m + 1)^.

83. a« - a* - a- + 1. 91 . (.r- - / - z^- - 4 t/^^^.

92. o2 4- 25 6^ _ 16 c2 - 9 d- - 10 a6 - 24 cd.

93. (1 + o^) + 2 (1 - a) (1 + a)2.

HIGHEST COMMON FACTOR. 81

X. HIGHEST COMMON FACTOR.

v6

107. The Degree of a rational and integral monomial
(§ 69) is the number of letters which are multiplied to-
gether to form its literal portion. ilj'TM-^'^

Thus, 2 a is of i\ve first degree ; 5 a6 of the second degree ; • .^/-t^M'^'
3 a^b^, being the same as 3 aabbb, is of the fifth degree ; etc.

The degree of a rational and integral monomial is equal
to the sum of the exponents of the letters involved in it.

Thus, a*bc^ is of the eighth degree.

108. A polynomial is said to be rational and integral
when each term is rational and integral ; as 2 a-b — 3 c + dl

The degree of a rational and integral polynomial is the
degree of its term of highest degree.

Thus, 2 a^b — 3c + d- is of the third degree.

109. A Prime Factor of an expression is a factor which
cannot be divided without a remainder\by any expression
except itself and unify.

Thus, the prime factors of 6a^(x^ —■ 1) are 2, 3, a,a,x-\- 1,
and X — 1.

110. The Highest Common Factor (H. C.F.) of two or
more expressions is the\product of all their common prime
factors.

It is evident from this definition that the highest common
factor of two or more expressions is the expression of high-
est degree (§ 108) which will divide each of them without a
remainder.

111. Two expressions are said to be prime to each other
when unity is their highest common factor.

82 ALGEBRA.

112. Required tlie H. C. F. of a*6V, a-6V, and a^bc*.
Resolving each expression into its prime factors, we have

a*bV = aaaabbccc,
a^b^(^ = aabbbccccc,
and a^C^ = aaabcccc.

Here the common prime factors are a, a, b, c, c, and c.

Whence, the H. C. F. = aabccc = a^bd^.

It will be observed, in the above result, that the exponent
of each letter is the loivest exponent ivith which it occurs in any
of the given expressions.

113. In determining the highest common factor of alge-
braic expressions, we may distinguish two cases.

114. Case I. WJien the expressions are monomials, oi
polynomials ivhich can be readily factored by inspection. .

1. Find the H. C. F. of 28 a'b^, 42 ab'c, and 98 a%*d?.
We have, 28 a%^ = 22 x 7 x a%^,

42 ab^c = 2 X 3 X 7 X ah^c,
and 98 a^hH- = 2 x 7^ x a%*cP.

By the rule of § 112, the H. C. F. = 2 x 7 x ah^ = 14 ab^, Ans.

EXAMPLES.
Find the highest common factor of :

2. 2a%5a^b\ 4. 45 a^ft", 120 aV.

3. 20a^y,15xf. 5. 182 oc^yz', S-ix'y'z.

6. 16 mV, 56 m*n% 88 m'li^

7. 36 a'bc^, 72 a^b\ 180 a6V.

8. 126 aV, 21 a'xh/, 147 a\v^z.

9. 140 m^n^x", 175 m^n% 105 m*n:^.
10. 117 a'b^c% 104 a^6V, 156 a^6V.

HIGHEST COMMON FACTOR. 83

11. Find the H. C. F. of

5 x'^y — 45 2t?y and 10 !x?y^ + 40 Qi?y^ — 210 xy"^.

We have, 6 od^y — 46 x'^y = 5 xhj (x^ — 9)

= 5x2y(x + 3)(x-3), (§ 99j

and 10 x3^2 ^. 40 x-y- - 210 x?/^ = 10 xy^ (x^ + 4 x -21)

= 2 X 5 X x?/2 (jc + 7) (X - 3). (§ 100)
By the rule of § 112, the H. C. F. is 5xy(x - 3), Ans.

12. Find the H. C. F. of

4 a^ — 4 a + 1, Sa^ — 1, and 2 arn — m — 2 an -\- n.

We have, 4a2-4a+l = (2a- 1)2, (§ 98) .

8 a3 - 1 = (2 a - 1) (4^a2 + 2 a + 1), (§ 103)
and 2 am — m — 2an -\- n = (2 a— 1) (m — n). (§ 93)

By the rule of § 112, the H. C. F. is 2 a - 1, Ans.

Find the highest common factor of :

13. 6 a^h" - 15 o'V', 12 a% + 21 a?W.

14. 68 (m + ny (rn — n)*, 85 (m + n)^ (m — w). N/^

• 15. a^-9/,x2-6a;i/+92/2. " ^U^

16. 3a3-21a2-a + 7, a'^-fGa-Gl. ^'

17. 2a^x + 4: aV' + 2 ax^, 3 a^a; + 3 ax*.

18. ms - 27, m^ - 11 m + 24.

19. ac + ad — be — bd, a^ — 6 ab -\- 5 b^.

20. a; + 4a^ + 4ar^4 + 44a; + 72a^.

21. 80 n^ - 5 n^, 20 n* + 5 nl

22. a2+&2_c2+2a6, a2-62_c2_,_2jc.

23. x' + 2x- 24, ar' - 14 a; + 40, x^-Sx + 16.

24. 9 a^ - 12 a + 4, 9 cr - 4, 18 a^ - 12 a^.

25. a^-6i»-27, x2 + 6a; + 9, ar^ + 27.

84 ^ ALGEBRA.

26. o? + 13 a^ 4- 40 a, a* - a' - 30 a", a' + 2 a* -15 al

27. m^ — 4 m, m^ + 9 wi^ — 22 m, 2 ??i^ — 4 wi^ — 3 m^ + 6 m.

28. x^-8f, x^-4:tf, x'-dxy + Uy'-.

29. 3 a^ - a-6 + 3 a& - 6^ 27 a^ -b'', 9a^-6ab+ h\

30. 27 a^ + 125, 9 ar= - 25, 9 ar + 30 a; + 25.

31. x^y-xY-20xf, 2 a^y^ +22 x^f^ 56 xy\ 3x'y-ASafy\

32. 16 m* — n*, 16 r*i* — 8 7?i-«-- + n*, 2 mx + 2 ??i?/ — ?ia; — ny.

33. a^ — iK^, a' — a^^ — ci3^ + ^, 3 a^ — 3 a^x + 5 ax*^ — 5 x^

115. Case II. When the expressions are polynomials which
cannot he readily factored by inspection.

The rule in Arithmetic for the H. C. F. of two numbers
is:

Divide the greater number by the less.

If there be a remainder, divide the divisor by it; and con-
tinue thus to make the remainder the divisor, and the preceding
divisor the dividend, until there is no remainder.

The last divisor is the H. C. F. required.

Thus, let it be required to find the H. C. F. of 169 and

546.

169)546(3
507

"39)169(4
156

~13)39(3
39

Then, 13 is the H. C. F. required.

116. We will now prove that a rule similar to that of
§ 115 holds for the H. C. F. of two algebraic expressions.

Let A and B be two polynomials, the degree of A (§ 1U8)
being not lower than that of B.

HIGHEST COMMON FACTOR. 85

Suppose that B is contained in A p times, with a i^femain-
der C; that C is contained in B q times, with a remainder
D ; and that D is contained in G r times, with no remainder.

To prove that D is the H. C. F. of A and B.

The operation of division is shown as follows :

B)A{p
pB

~C)B(q
qC

D)C{r
rD

0

We will first prove that Z) is a common factor of A and B.

Since the minuend is equal to the subtrahend plus the
remainder (§ 35), Ave have

A=pB+C, (1)

B = qC + D, (2)

and C = rD.

Substituting the value of C in (2), we obtain

B = qrD + D = D (qr + 1). (3)

Substituting the values of B and C in (1), we have

A =pD (qr + 1) +rD = D (pqr +p + r). (4)

From (3) and (4), Z) is a common factor of A and B.
We will next prove that every common factor of A and B
is a factor of D.

Let F be any common factor of A and B ; and let

A = mF and B = nF.
From the operation of division, we have

C=A-pB, (5)

and D = B~ qC. (6)

86 ALGEBRA.

Substituting the values of A and B in (5), we have

C=mF—pnF.
Substituting the values of B and C in (6), we have
D = nF — q (mF — pnF) =F(n — qm-\- pqn).
Whence, i^ is a factor of D.

Then, since every common factor of A and jB is a factor
of D, and since D is itself a common factor of A and B, it
follows that D is the highest common factor of A and B.

117. Hence, to find the H. C. F. of two polynomials,
A and B, of which the degree of A is not lower than that
of JB,

Divide Aby B.

If there be a remainder, divide the divisor by it; and con-
tinue thus to make the remainder the divisor, and the preceding
divisor the dividend, until there is no remainder.

The last divisor is the H. C. F. required.

Note 1. Each division should be continued until the remainder
is of a lower degree than the divisor.

Note 2. It is of the greatest importance to arrange the given
polynomials in the same order of powers of some common letter
(§ 33), and also to arrange each remainder in the same order.

1. Find the H. C. F. of

6a:2_ 13 a;- 5 and 18 0:^-51x2 +13 a; + 5.

6x2 _ i3x _ 5)18 a;3 - 51 a;2 + 13a; + 5(3a; - 2
18x8- 39x2 -15x

- 12 x2 + 28 X
-12x2 + 26x + 10

2x- 5)6x2 -13x-5(3x + l
6x2- 15 X

2x

2x-6
Whence, 2 x — 6 is the H. C. F. required.

HIGHEST COMMON FACTOR. 87

Note 3. If the terms of one of the given expressions have a
common factor which is not a common factor of the terms of the other,
it may be removed ; for it can evidently form no part of the highest
common factor. In like manner, we may divide any remainder by a
factor which is not a factor of the preceding divisor.

2. Find the H. C. F. of

6 a^ — 25 a^ + 14 X and 6 aoi? + 11 ax — 10 a.

In accordance with Note 3, we remove the factor x from the first
expression, and the factor a from the second.

6 a;2 _ 25x + 14)6 a;2 + 11 X - 10(1
Qx^-2bx + 14
36 « - 24
We divide this remainder by 12 (Note 3) .

3x -2)6x2-25a; + 14(2x-7

-21x

- 21 X + 14

Whence, 3 x — 2 is the H. C. F. required.

Note 4. If the given expressions have a common factor which
can be seen by inspection, remove it, and find the H. C. F. of the
resulting expressions. The result, multiplied by the common factor,
will be the H. C. F. of the given expressions.

3. Find the H. C. F. of

2a^-2>a?h-2 aW and 2a? + l a?h + ^ ah\

In accordance with Note 4, we remove the common factor c, and
find the H. C.F. of 2 a^ - 3 a& - 2 62 and 2 a^ + 7 a& + 3 62.

2a2-3a6-262)2a2+ 7a6 + 362(l
2 a2 - 3 a6 - 2 62
5 6)10a6+_562
2a + h
2 a + 6)2 o2 - 3 a6 - 2 62(a - 2 6
2 a» + a6
-4a6
- 4 a6 - 2 62

Multiplying 2 a + 6 by a, the required H. C. F. is a(2 a + 6).

88 ALGEBRA.

Note 5. If the first term of the dividend, or of any remainder, is
not divisible by the first term of the divisor, it may be made so by
multiplying the dividend or remainder by any term which is not a
factor of the divisor.

Note 6. If the first term of any remainder is negative, the sign
of each term of the remainder may be changed.

4. Find the H. C. F. of

2 a!^ - 3aj2 + 2a; - 8 and 3a.-3 - 7 ar + 4a; - 4

3x3- 7x2 + 4x-4
2
2x3 _ 3x2 + 2x - 8)6x3- 14x-i + 8x- 8(3
6x3- 9x2 + 6x-24

— 5 x'^ + 2 X + 16

2x3_ 3x2+ 2x- 8-
5
6x2 -2x- 16)10x3- 15x2+ lOx- 40(2x
10x3- 4x2- 32x

-11x2+ 42 X- 40
6
-65x2 + 210x-200(-ll
-55x2+ 22 x+ 176

188^188x-376

X- 2

X- 2)5x2- 2x-16(5x + 8
6x2 -lOx
.8x
8x- 16

Whence, x — 2 is the H. C. F. required.

In the above example, we multiply 3x3 — 7 x2 + 4x — 4 by 2 in
order to make its first term divisible by 2 x3.

"We change the sign of each term of the first remainder (Note 6),
and multiply 2 x3 — 3 x2 + 2 x — 8 by 5 to make its first term divisible
by 5 x2.

We multiply the remainder — 11 x2 + 42x — 40 by 5 to make its
first term divisible by 5 x2.

HIGHEST COMMON FACTOR. 89

EXAMPLES. i

Find the H. C. F. of :

5. 2 a-2 - 5 a; + 3, 2 ar — 7 a; + 5.

6. 2a- + 7a + 6, 6 a- + 11 a + 3.

7. 4 x~ + 13 a; + 10, 6 ar + 5 ic — 14.

8. x' + ox-24., af5 + 4a^-26a; + 15.

9. 3 m^ + m — 2, 4 m^ + 2 m^ — m + 1.
10. 18 a^ + 9 ah - 5 b\ 24 a- - 29 A& + 7 h\

U. 12 a^ - 5 a^a;- 11 aa^-^^ 6 a;^, 15 a^ + 11 a^a; - 8 oa^ - 4 a^.

12. 4 ar' - 12 x' + bx, 2 x^ + o? -1 x^ -20 x.

13. 3 ar^ + 13 ar> + 12 a^?/2, ^a?y -22x^f -^y\

14. 4a* - 11 a^ + 5a + 12, 6a«- 11 a^ + 13 a^' - 4 a^.

15. 2 ?»* + 5 m% — 2 mhv' + 3 m7i^, 6 m^n — 7 m^?i- + 5 m'n? — 2 7i*.

16. 3a^-4a;-4, 3a;* - 7ar' + 6a;2- 9a; + 2.

17. 3a* + 5a3 + 12a2 + 8, 6a*+ lOa^ + 19a- - 10a -4.

18. 2 m^ — 3 m^.r — 8 ma;^ — 3 .x^, .

3 m* — 7 7n^a; — 5 ??i^a;^ — ma;^ — 6 a?*.

19. 2 a* - a^ - 4aM- 3 a, 4 a* - 6 «■'' + a^ + 4 a - 3.

20. vi^ + 8 m^, m^ — 2 m* — 15 m^ — 14 w?.

21. 4a*-22a"6 + 6a262-f-20a6^ 9a36-42a2&2_18a&»+156*.

22. 4 .r^ + 9 .T - 9, 2a;* + llar' + 14aj2_5^_g_

23. 3 a*- 6 a3+ 4 a-+ 4 a - 4, 3 a'+ 3 a*- 11 a?- 2 a' + 6 a,

24. a3 + 2a2-2a + 24, a' + 2a=^ - lla^ - 6a + 24.

25. 2a;*-3ar'y + 3.^•22/2-3a;?/=^ + 2/*,

2 a;* + .^'^^/ — 3 xry^-\- 5 a;^^ — 2 y*.
m 2x* + ar'-9a.-2 + a; + l, 2a;''-9a;3 + 12a;2_3^_2.

27. 2 3.-3-7 3.-' + 7 a; -2, a;^ - 3 a;* + Sa;^ - 4 a; + 4.

28. a^x — a*x' — aV — a^x* — 2 ax^,

a^x + 3 a*a;^ — a^a;^ — 4 a^a;* — aa;'.

90 ALGEBRA.

118. The H. C. F. of three expressions may be found as
follows :

Let A, B, and C be the expressions.

Let G be the H. C. F. of A and B ; then, every common
factor of G and (7 is a common factor of A, B, and G.

But since every common factor of two expressions exactly
divides their highest common factor (§ 116), every common
factor of A, B, and G is also a common factor of G and C.

Whence, the highest common factor of G and G is the
highest common factor of A, B, and C

Hence, to find the H. C. F. of three expressions, find the
H. C. F. of tivo of them, and then of this result and the third
expression.

We proceed in a similar manner to find the H. C. F. of
any number of expressions.

1. Find the H. G. F. of

a^_7a; + 6, a^ + Sx" - 16aj + 12, and o^-bx' + lx-^.
TheH. C.F. of a;^ - 7a; + 6 and x^ + Zx"- 16a; + 12 isa;'--3x + 2.
The H. C. F. of x2 - 3 X + 2 and x^-Sx^ + Tx-Sisx-l, Ans.

EXAMPLES.
Find the H. C. F. of:

2. 2 x2 - 17 a; + 36, 4 x^ - 12 a; - 27, 6 a^ - 31 x + 18.

3. 8a2 + 22a + 5, 12 a^- 13 a -4, 20a2 + 29a + 6.

'4. 15 m^ - 4 7?i - 32, 18 m" + 3 ?7i - 28, 21 m^ + 25 m - 4.

5. 5a2 + 23a&-10&2, 5 a^ + 33 a^^ + 46 aft^ _ 24 &',

5 a^ + 38 a^h + 59 ab- - 30 h\

6. af'+a;2- 14 a; -24, a? -^o? -Qx -^?,, ct? -\- 4.0? + x - Q>.

7. a3 - a2 _ 5 a - 3, a^ + 2 a^ - a - 2, a^-2a^-2a + l.

8. 2 m^ + 9 ??i2 _ 6 7>i - 5, 3 m^ + 10 m"" - 23 m + 10,

6 m^ — 7 m' — m + 2.

9. 2a^~x'y-2^x^f-\-3Qf, 2 x^ - 5x-y ~ 37 xy- + 60 f,

2a^-19x'^y + 54:xy^ - A5f.

LOWEST COMMON MULTIPLE. 91

XI. LOWEST COMMON MULTIPLE.

119. A Common Multiple of two or more expressions is
an expression wliich can be divided by each of them with-
out a remainder.

120. The Lowest Common Multiple (L. C. M.) of two or

more expressions is the product of all their different prime
factors (§ 109), each taken the greatest number of times
that it occurs as a factor in any one of the expressions.

121. Required the L. C. M. of a^b% a¥d% and &Vd*.
Here, the different prime factors are a, b, c, and d; a

occurs twice as a factor in a-b^c; b five times as a factor in
ab^d-; c three times as a factor in 6Vd*; and d four times as
a factor in b-cM\

Whence, the required L. C. M. is a-b^c^d* (§ 120).

It will be observed, in the above result, that tJie exponent
of each letter is the highest exponent with which it occurs in
any one of the given expressions.

122. It is evident from the definition of § 120 that the
lowest common multiple of two or more expressions is the
expression of Imvest degree (§ 108) which can be divided by
each of them without a remainder.

123. If two expressions are prime to each other (§ 111),
their product is their lowest common multiple.

124. In determining the lowest common multiple of
algebraic expressions, we may distinguish two cases.

125. Case I. When the expressions are monomials, or
polynomials which can be readily factored by inspection.

92 ALGEBRA.

1. Find the L. C. M. of 2Sa'h% bUc\ and 63 cU
We have, 28 a'6- = 2'^ x 7 x a^h",

54 6c3 = 2 X 33 X &c3,
and 63 c2d = 32 x 7 x cH.

By the rule of § 121, the L. C. M. = 22 x 33 x 7 x a*b-cH

= 756 a'^b'2c% Ans.

EXAMPLES.
Find the lowest common multiple of:

2. 5a&^ 7a-b-. 6. 55 xy, 70 yz, 77 zx.

. 3. 12 xf, 54. yz\ 7. 50 a'b% 60 a'b% 75 a'b'.

4. 24. m^, 45 nl 8. 15 xV, 21 fz, 33a;V.

5. 72 a% 96 &V. 9. 20 ab% 27 &V, 90c*d2.

10. oQin^nx, 40 mn^y*, 4Sn^3?y.

11. 56o?bc% Ua%H', 12Q aJc-d\

12. Find the L. C. M. of .'B2 + a;-6, a^-4a; + 4, and
a^ — 9 a;.

We have x2 + a; _ g ^ ^^ _^ 3) (^3. _ 2)^ (^§ 100)

x2-4a; + 4 =(x-2)2, (§08)

and x3-9a; = x(x + 3)(x-3). (§99)

By the rule of § 121, the L. C. M. = x{x - 2y-{x + 3) (x - 3), Ans.

Find the lowest common multiple of :

13. a" - b', o? + 2 ab + b\

14. m^ + mn, mn — n^.

15. x'-d, x^ + lOx + 21.

16. x' - 18 x^ + 81 x", x^ - 13 x^ + 36 a;.

17. a^-Sab + 2b^, ac + ad-bc-bd.

18. a^ + 2 a.c + ar^, a^ + x^.

19. 1 - 8 ar'', 1 + 9 a; - 22 x".

LOWEST COMMON MULTIPLE. 93

20. nv^ + 13 m^n + 40 7nn-, mhi — mn^ — 30 n^.

21. 4 ar - 25, 2 ar'' - 5 or' - 4 a; + 10.

22. x-3 + 3 aa^ - 18 a% ax^ + 15 a^a; + 54 a^

23. 4d'-2ab, 4 a6 + 2 6^ 4 a- - b\

24. 6 a;- + 10 xy, 9 a;y - 15 7/, 36 ar*?/ - 100 xf.

25. 4 m- — 8 m + 4, 6 nr + 12 vji + G, m^ — 1.

26. a^ - 12 a + 35, a'^ + 2 a - 63, a" -3a- 108.

27. a;* — 4 aa:^ + 4 a-a^, a.-^ + 4 ax + 4 ct^, aar' — 4 a^x.

28. 3a^-6a;-72, 4a^ + 8a;-192, 2a^ -24:X-\-72.

29. a."^?/ — xy^, x^ — y^, x^ — 2 xy + y^.

30. a."^ + 2/^ — 2^ — 2 a;?/, x^ — y- — z^ — 2 yz.

31. 16 7?i^ — 9 «^, 8 a6-77i — 6 atrn, 16 m^ — 24 mn + 9 w^.

32. a^ — a, a^ — 9ar — 10 a, a* — a^ -\- a? — a.

33. ar + 4 xy + 4 ?/^, x^ + xy — 2 y-, a;^ + 8 ?/^.

34. 2a='-2a2-4a, 3 a*- 6a='- Oa^, 4a^ + 20a* + IGa^.
"35. 27aT^-8, 9a!2-4, 9ar-12a: + 4.

36. 4:0^-4.^1% 6x+67n, Sx' + Sm'', 9x-9 m.

37. x* - y*, x* + 2 xY' + y\ x' - 2 ar^y/^ + i/.

38. a' + b^, a? - b\ (a' + b^f - a%\

39. a"— 11 aa; + 18 a^, a^— 5 aa; — 14 a;^, a*— 8a^x^+ 16a;^

40. m' — 71", m^ — m^?i — fnv? + w^, m^ + mhx — triy^ — it?.

41. a2+ 62_ c2+ 2 a&, a^- 6^- c^- 2 6c, a-- 6^+ c^- 2 ac.

126. Case II. Wlien the expressions are polynomials
which cannot be readily fctctored by inspection.

Let A and B be any two expressions.
Let F be their H. C. F., and M their L. C. M. ; and sup-
pose that A = aF, and B = bF.

94 ALGEBRA.

Then, AxB = ahF\ (1)

Since F is the H. C. F. of A and B, a and h have no com-
mon factors ; whence, the L. C. M. of aF and hF is ahF.

That is, M= ahF.

Multiplying each of these equals by F, we have

FxM= abF\ (2)

From (1) and (2), AxB = FxM. (§ 9, 4)

That is, the product of two expressions is equal to the prod-
uct of their H. C. F. and L. C. M.

Therefore, to find the L. C. M. of two expressions.
Divide their product by their highest common factor ; or.
Divide one of the expressions by their highest common fac-
tor, and multiply the quotient by the other expression.

1. Find the L. C. M. of

6ic--17a! + 12 and 12a^-4a;-21.

6ic2 _ 17 x+ 12)12x2- 4x-21(2
12x2-34x + 24

15)30x-

-45

17x

+ 12(3x-

'2a;-

- 3)6x2-

-4

6x2-

9x

-

8x

-

8x

+ 12

Then the H. C. F. of the expressions is 2 x — 3.

Dividing 6 x2 — 17 x + 12 by 2 x — 3, the quotient is 3 x — 4.

Whence, the L. C. M. = (3 x - 4) (12 x2 - 4x - 21), Ayis.

EXAMPLES.
Find the L. CM. of:

2. 2x'-^x-^b,2x'^-19x + 45.

3. 3 a^ - 13 a + 4, 3 a^ + 14 a - 5.

4. 6 a^ + 25 ab + 24 b\ 12 a^ + 16 a6 - 3 h\

5. 6 x^ + 11 x-y - 2 xy\ 8 x^y + 21 xy"" + 10 f.

LOWEST COMMON MULTIPLE. 95

6. 12 m^ - 21 m - 45, 4 m^ - 11 m^ - 6 m + 9.

7. 2 a^- 5 a' -IS a -9, 3 a" - 14 a- - a + 6.

8. 2 a'x + a-.r + 2 ax^ + 3 x\ 2 aU- + o arxr + 2 aar^ - a;*.

9. 2a'-5ab + 3 h\ a' + a'b - 5 arh- + 2 aZ^^ + h\

10. 6a^-7x-2 + 5a;-2,4x-*-5ar' + 4aj-3.

11. 2a3-5a'^ + a + 2, 4a3-9a-4.

12. 3 m* — 7 //i^n + 4 wi/i", G wi^/i — 4 mhi^ — 14 tnn^ — 4 w*.

13. a^ + 2 a^ - 5 a^ + 12 or, 3 a" + H a' - 6 a^ ~ 7 a^ + 4a-.

14. •Sx^-2x^-12x^-x + Q>,3x' + lx^ + Qx'-2x~L

127. The L. C. M. of three expressions may be found as
follows :

Let A, B, and C be the expressions.

Let M be the L. C. M. of A and B ; then, every common
multiple of 31 and C is a common multiple of A, B, and C.

But since every common multiple of two expressions is
exactly divisible by their lowest common multiple, every
common multiple of A, B, and C is also a common multiple
of 3/ and C.

Whence, the lowest common multiple of M and C is the
lowest common multiple of A, B, and C.

Hence, to find the L. C. M. of three expressions, find the
L. C. M. of two of them, and then of this result and the third
expression.

We proceed in a similar manner to find the L. C. M. of
any number of expressions.

EXAMPLES.

Find the L. C. M. of:

1. 2x- + x- 15, 2 .T- + 7 re + 3, 2 a;2 + 9 a; + 9.

2. 3 a- + a — 2, 6 cr + 11 a + 5, 9 a- + 5 a — 4.

3. 2m^-5m-\-2,3 m- - 10 m + 8, 4 m^ + 10 m - 6.
t.^ 23^-5x^-3x, 4:X*-llaf-3x^, 6a^-x'-2x'.

5. a^-2a^-5a + 6,a^-3a'-a-{-3,a^-\-4:a^ + a-6.

96 ALGEBRA.

a

XII. FRACTIONS.

128. The quotient of a divided by h is written - (§ 3).

The expression - is called a Fraction ; the dividend a is

called the numerator, and the divisor h the denominator.

The numerator and denominator are called the terms of
the fraction.

129. Let ^ = x. (1)

Then since the dividend is the product of the divisor and

quotient (§ 54), we have

a = hx.

Multiplying each of these equals by c (§ 9, 1),

ac = hex.

Kegarding ac as the dividend, he as the divisor, and x as
the quotient, this may be written

f^ = x. (2)

From (1) and (2), ^ = ^- (§9,4)

he h

That is, if the terms of a fraction he hoth multiplied, or both
divided, hy the same expression, the value of the fraction is not
altered.

130. By the Law of Signs in Division (§ 55),

+ (x__— a_ +a_ —a

T&~^~ ^^~ +h
That is, if the signs of both terms of a fraction be changed,
the sign before the fraction is not changed ; hut if the sign of
either one be changed, the sign before the fraction is changed.

FRACTIONS. 97

If either term is a polynomial, care must be taken, on
changing its sign, to change the sign of each of its terms.

Thus, the fraction ^ ~ , by changing the signs of both
c — d , _

numerator and denominator, can be written — (§ 41).

d — c

131. It follows from §§49 and 130 that

If either term of a fraction is the indicated product of two
or more expressions, the signs of any even number of them
may be changed ivithout changing the sign before the fraction ;
but if the signs of any odd number of them be changed, the
sign before the fraction is changed. '

Thus, the fraction ^ may be written

(c - d) (e -/)

a — b b — a b — a .

etc.

{d - c) (/- e)' id - c) (e -/)' (d - c) (/- e)'

REDUCTION OF FRACTIONS.

132. To Reduce a Fraction to its Lowest Terms.

A fraction is said to be in its lowest terms when its numer-
ator and denominator are prime to each other (§ 111).

133. Case I. When the numerator ayid denominator can
be readily factored by inspection.

By § 129, dividing both terms of a fraction by the same
expression, or cancelling common factors in the numerator
and denominator, does not alter the value of the fraction.

We then have the following rule :

Resolve both numerator and denominator into their factors,
and cancel all that are common to both.

1. Reduce — to its lowest terms.

40 a-VcZ

We have, gjo^j^c ^ 23 x 3 x g^&^c _

40 a-bhl 23 X 5 X a-b'^d

98 ALGEBRA.

Cancelling the common factor 2^ x a-ft^, we obtain

40 a^b^d 5 d

2. Reduce to its lowest terms.

x^ — 2x — 3

We have, -^-^^ = (x - 3)(x^ + Sx + 9) ^^^^ ^^^^

a;2 _ 2 X — 3 (x — 3) (« + 1)

^^+Sx + 9 ^^^^
x+1

Note. If all the factors of the numerator be cancelled, unity re-

mains to form a numerator : thus, -^ = -—•

x^y^ x-^y

If all the factors of the denominator be cancelled, the division is

exact.

EXAMPLES.

Reduce each of the following to its lowest terms :

g aW_ g 32 ab* g ISab^

ah'& 72 a%' 108 a'b'c'

7m*n^p m 56 aSin?n^ jq 60 m^n^x^

2m^nY' ' 84 a*mV* ' d^jm^i^x'

54ccy f. 120 .tV^^ jj 126a^6V

^' 457?' 15^"^' * ' 98a6»V*

12 3ft'&-6aT 17, m3-m2-56m

4 a'&2 _ 8 a&3 m" + wi'' - 42 m^

6a;^y + 8a^y^ ,« a^ + y"

^'*- 15a;V + 20a^/ 2 or'?/ - 2 a^^/^ + 2 a^'

a2^7q^l0 64a^ + 112a''.r + 49aa^

^*- a2_|_4a-5' ' 64 a^a; - 49 a;''

jg a;^ — 8a^ + 12a; £0 ^ — 14ma; + 45 m^

a^_12ic + 36 a;'- 2 ma; -15 m'

'' + 20a6 + 46^ 21 «'-8

25a2-462 * ■ a='-2a« + a-2'

FRACTIONS. 99

22 47/1^-10771=^ -6m + 15 (a^-9)(a^ + 5a + C)

' x^ — y^ — z^ -{-2yz' (a — ciy — (b — c)^

24 27a^ + 646^ 2« 12a^ + 8a;- -3a;-2

9a- + 24a& + 1662' ' ISx^-ga;^- 8a; + 4*

no -r, T ax~bx — ay + by^ . , ,

28. Reduce 5-5 ^^ ^ to its lowest terms.

&^ — a-

Wehave ax - to - ay + 6^/ _ (a - &)(x- y)

we nave, ^, _ ^^ - (j, + „)(6 _ «)• C§^ 9^, JJ)

Changing the signs of the factors of the numerator (§ 131), we have

ax — bx — ay + by _(b — a)(y — x) _y — X

62 _ a'2 ~ (6 + a)(6-a) ~ F+a' "*'

Reduce eacli of the following to its lowest terms :

QQ 9 — m^ 00 2 ac — 2 6c — ad + bd

■ m^ — 7 771 + 12

30. ^^^-^^ . 33.

4a;2_28a; + 49

31^ g^-7a;y+6/^ g^

y^ — x^ ' (b — a)- — (?

134. Case II. When the numerator and denominator can-
not be readily factored by inspection.

Since the H. C. F. of two expressions is the product of all
their common prime factors (§ 110), we have the following
rule:

Divide both numerator and denominator by their highest
common factor.

1. Reduce - — - — '- — ^^— to its lowest terms.
6 a^ - a - 12

By the rule of § 117, we find the H. C. F. of 2a2 - 5a + 3 and
6a2- a- 12 to be 2a -3.

d'

-4c2

1-

-11a-

4-18a2

8a«-

-1

a"-

-(& +

cy

100 ALGEBRA.

Dividing 20"^ — 5a + 3 by 2a — 3, the quotient is a — 1.
Dividing Q a^ — a — 12 by 2 a — 3, the quotient is 3a + 4.

Wheuce, 2a^-ba + S ^ a^^ ^^^^

6 a- -a -12 3 a + 4

EXAMPLES.
Reduce each of the following to its lowest terms :

- Sar + Uab + Sb^
4 a' + 15 ab-4tb^'

' 3a)3-14ar'-lliB-2*
2a^ + 9ci'-2a-3

2.

a^_3a;-18

5x^-23^- -42

3.

2a2 + a-10

4a2 + 8a-5

4.

2 a.*^ — xy — 15 ?/^

2a32-15x?/ + 27y»

6.

6m2-13m4-6

9m2 + 6m-8

R

a,^ + .3a;-10

9.

6a^ + 23a2-22a + 3

.^ m^ + ffl'^ + w + 6
m^ + 6 m^ + 6 m — 4

11 «^ + 2 ft^a; — 2 ax^ — a^
"■ ar' + 2a^-14ic + 5 ' a" - 3 a-x -2 ax^ + 4 a^'

135. To Reduce a Fraction to an Integral or Mixed Ex-
pression.

An Integral Expression is an expression which has no
fractional part ; as 2 xy, or a + b.

An integral expression may be considered as a fractioii
whose denominator is 1 ; thus, a + 6 is the same as -—■ —

A Mixed Expression is an expression which has both
integral and fractional parts j as a + -, or x + -

136. We have by § 30,

ax(- + ~)^ax- + ax-=b + c. (§9,3)
\a aj a a ■ ^

FRACTIONS. 101

Regarding & + c as the dividend, a as the divisor, and

be

- + - as the quotient (§ 54), this may be written

b + c_ be
a a a

137. A fraction may be reduced to an integral or mixed
expression by the operation of division, if the degree (§ 108)
of the numerator is equal to, or greater than, that oi the
denominator.

1. Reduce ^t^ ^^— to a mixed expression.

By§136, =: — + ---- = 2x + 5--,Ans.

2. Reduce — ■—■ to a mixed expression.

4a^ + 3 ^

4a;2 + 3)12x» -8x'2 + 4:X- 5(3a; - 2
12x3 +9a;

-8x2_5a;
•-8x2 -6

-5x+ 1

A remainder of lower degree than the divisor may be written over
the divisor in the form of a fraction, and the result added to the
quotient.

Thus, 12a;^-8x2 + 4x-5^3^_ -5X+1.
4 x2 + 3 4 x2 -j. 3

If the first term of the numerator is negative, it is usual to change
the sign of each term of the numerator, at the same time changing the
sign before the fraction (§ 130).

Thus, 12x3-8x2 + 4x-5 ^ 3^ _ 2 _ 5x - 1 ^„^^

4x2 + 3 4x2 + 3

EXAMPLES.
Reduce each of the following to a mixed expression :

3 12a^-16a; + T ^ 15 a^ + 6a^ - 3a - 8,

4a; * . ■ 3 a

102 ALGEBRA.

2ic + 3 ' x — y ' a-\-h

3a + 4 * ■ 4m2 + l

ic + y

2m-

■ 5n

2a?-

■Sx"

-6

x"-

- X —

1

12 a^.

-5a

-5

10. ^a^-^^-o. 14

11. ±±" -""-'-. 15.

4a — 1

18 g'^ - 3 g'^ + 38
3a2-4a + 5

a*+&^

•Q 8 a^+ 16 a^- 10x^-28 a; + 11
2iif + x--8

138. To Reduce a Mixed Expression to a Fraction.

The process being the reverse of that of § 137, we have
the following rule :

Multiply the integral part by the denominator.

Add the numerator to the product when the sign before the
fraction is +, and subtract it when the sigri is — ; and write
the result over the denominator.

1. Reduce -^-^ 1- a; — 2 to a fractional form.

2a;-3

We have, -^+A + ^ „ 2 ^x + 5 +(x - 2)(2x - 3)
' 2»-3 2x-3

_a; + 5 + 2a;'^-7a; + 6
2a;-3

2a;-3

If the numerator is a polynomial, it is convenient to en-
close it in a pai^enthesis when the sign before the fraction
is — .

FRACTIONS, 103

2. Reduce a — h — - to a fractional form.

a + h

0 + 6 a + 6

_ai - V^ - a^ ■\. ah + l^
a + 6
ab

a + &

, ^ra&

EXAMPLES.
Reduce each of the following to a fractional form :

3. a-4 + 2±2. „_ ^ + 23,-^+|i^.

«■ l^i+l- la. 4m'-9 + 6^2i2»-3i.

^ 2m + 6

*■ ^" + ^-2^- 13. 2a^ + 3a-^''(«-^).

2a — 1

CO o 11 a^ + 7 , n^ — 11?

6- 3^-2 -^. 14. ^ I^ + a_56.

'^^' 4a — 36

7 ^_3a-& a-3 + 2/3

2 w^ o w

8. m2-mw + %8-— - — 16. aS+a^ft -ha&^+fts^ jf o

w + w • . . i . ^_^

9. |^+|^_l. 17. (^-l)^-(a^-a; + l).

2a — 5a; or^ + aj + l

10. do? + 4 + — — . 18. m4-3« 5—- -——

3 a; — 4 m^— 3 m« -j-9 w^

139. To Reduce Fractioiis to their Lowest Common De-
nominator.
To reduce fractions to their Lowest Common Denominator

(L. C. D.) is to express them as equivalent fractions, having
for their common denominator the lowest common multiple
of the given denominators.

104 ALGEBRA.

Let it be required to reduce i^, ?^^, and ^^^^ to

their lowest common denominator.

The L. C. M. of 3 a% 2 ah\ and 4 a% is 12 a%'' (§ 125).

By § 129, if both terms of a fraction be multiplied by the
same expression, the value of the fraction is not altered.

Multiplying both terms of — ^—- by 4 ah, both terms of

3 a-h

- — - by 6 a^, and both terms of '- — - by 3 h, we have

16 ahcd 18 (j^iirx ■, 15 hnv

■ , ■ , and ^•

It will be seen that the terms of each fraction are multi-
plied by an expression which is obtained by dividing the
L. C. D. by its own denominator ; whence the following rule :

Find the lowest common mrdti])le of the given denominators.

Divide this by each denominator separately, midtijiily the
corresponding numerators by the quotients, and ivrite the
residts over the common denominator.

Before applying the rule, each fraction should be reduced
to its lowest terms.

140. 1. Eeduce -i-^ and -^—^ to their lowest

a- — 4 a^ — 5a-f6

common denominator.

"We have, a^ - 4 = (a + 2)(a -2), and a- - 5 a + 6 = (a -2)(a-3).

Then the L. C. D. is (a + 2) (a - 2) (a - 3). (§ 125)

Dividing the L. C. D. by (a + 2) (a — 2), the quotient is a — o ;
and dividing it by (a — 2) (a — 3), the quotient is a + 2.

Multiplying 4 a by a — 3, the product is 4 a(a — 3); and multiply-
ing 3 a by a + 2, the product is 3 a (a + 2).

Then the required fractions are

4 a(a - 3) _ , Sa(a + 2)

and ""^"^"^ , Ans.

(a + 2)(a-2)(a-3) (a + 2)(a - 2)(a -3)

FRACTIONS. 105

EXAMPLES.
Reduce the following to their lowest common denominator.
- 5xy Sicz 4:yz « 3 a; 5 a;

6 ' 14 ' 21

2 m^n 3 mn^ 5 m^r?

7.

^ 2a — 5o 4a + 36 g

9a-b ' 12a(? '

c Taz^ 9by 8ca^

9.

Gar' + S

)x 9a;2.

-1

ax

V

cx^y

x+y {

:^+2/)^'

{x+yf

2a

4 62

a^-h^'

a3+63

3

6

r»

9

8a^^ 10 a;^^' 152// ' « + l' « - l' a^ + l*

10.

11.
12.

2 a; a?

3a^-12x^ x'-Qx + s' a^-S'

x-\-y a — b

ax — bx—ay + by a? — 2xy -\- y^

a + 5 a + 3 a — 2

a2 - a - 6 a" + 7 a + 10 a^ + 2 a - 15

ADDITION AND SUBTRACTION OF FRACTIONS.

141. We have by § 136,

b , c 6 4-c

- + - = •

a a a

In like manner, r _ £. = ~^.
a a a

Whence the following rule :

To add or subtract fractions, reduce them, if necessary, to
equivalent fractions having the lowest common denominator.

Add or subtract the numerator of each resulting fraction,
according as the sign before the fraction is + or —, and write
the result over the lowest common denominator.

The final result should be reduced to its lowest terms.

106 ALGEBRA.

142. 1. Simplify in^+ll^«^.
4 a-b 6 ab^

The L. C. D. is 12 a^fts.

Multiplying the terms of the first fraction by 3 b^, and the terms
of the second by 2 a, we have

4a + 3 l-6b^_12ab^ + 9b'^ 2a~12ab^
^a'^b Qab^ 12 a^b^ 12 a-b^

_l2ab'^ + 9b^ + 2a- 12 ab^ ^ 9 6'^ + 2 g ^ ^

If a fraction whose numerator is a polynomial is preceded
by a — sign, it is convenient to enclose the numerator in a
parenthesis preceded by a — sign, as shown in Ex. 2.

If this is not done, care must be taken to change the sign
of each term of the numerator before combining it with the
other numerators.

2. Simplify g-^ j^.

The L. C. D. is 42.

Whence 5x-4y 7 x -2y ^S5x -28y 2lx-6y
'6 14 42 42

_35x-28y-(21x-6y)
~ 42

_35x-28y-21x + 6y
~ 42

_14x-22y_7x-lly

EXAMPLES.
Simplify the following :

3 5a-6 Sa + 7 - 3a; + 4 2a; + 5

8 12 * 12 16 ■

A _4 6_^ g CT — 4a; _ 7 x — 6a^

' 3xy^ 5o(^y ' 6ax^ 9a^x

FRACTIONS. 107

- x — 3m ix + m q 2a— 9 3a— 5 4ct+7

* 24m 32cc ' 7 14 28 '

g 2a— & 2b — c 2c-a ,q x+1 3 a;— 4 5cc+7
a6 6c ca 2x hv? Soc^

5a + l 26 + 3 7c-4

11.

6 a 8 6 12 c

12 ^^ — y,^^ — ^y 6ar' + 2?/^

5 a: 10?/ 15 xy

13 6a; + l 5a; — 2 8a; — 3 7 x + A

3 6 9 ~ 12 '

j^ 3a + 4 4a- 3 _ 5a-|-2 _ 6a- 1

,g 2a — 36 3a + 6 4a — 56 5a + 76
9 18 27 36 '

16. Simplify ^ ^

a;^ + a; af — x

We have, x'^ + a; = x(x + 1), and x^ — x = x(x — 1).
Then the L. CD. is x(x + l)(x - 1), or x(x2 - 1).
Multiplying the terms of the first fraction by x — 1, and the tern
of the second by x + 1, we have

1 1 x-1 x+1

X2 + X X2-X X(X2-1) X(X2-1)

^X-1-(X + 1)^X-1-X-1^ -2 ^^g

X(X2-1) X(X2-1) X(X2-1)'

By changing the sign of the numerator, at the same time chang
ing the sign before the fraction (§ 130), we may write the answer
2
x(x2 - 1)"

Or, by changing the sign of the numerator, and of the factor x^ — 1

2
of the denominator (§ 131), we may write it

108 ALGEBRA.

Wehave, a2_3a+2 = (a-l)(a-2),a2-4a + 3=(a-l)(a-3) and
a2_6a + 6=(a-2)(a-3;.

Then the L. C. D. is (a - 1) (a - 2) (a - 3).
Whence, — — +

a2-3a + 2 a'^-'ia + 'S a^-6a + 6

a-3 2(a-2) a-1

~(a-l)(a-2)(a-3) (a-l)(a-2)(a-3) (a-l)(a-2)(a-3)

^a-3-2(a-2)+a-l_a-3-2a + 4 + a-l^Q ^„S
(a-l)(a-2)(a-3) (a - l)(a - 2)(a - 3)

Simplify the following :

18 2 1 23 wi + w , m — w

■ 3a + 5 4a-7'

19. _J?!^ 1_. 24.

m — 1 m + 1

20. — ^- 25.

2a; + l 5a;-6

21. _^L_ + _^ — 26.
a+b a—b

22 ^^ _ 2 CT^ — 6 g — 3 07

"* • a + 4 a'-3a-2S ' 4.a'-9b^ (2a + 3by

1 1

m -

r

- n

m +

w

1-

X

l + a;

1 +

X

1-a;

4a2

+ 1
-1

2a

-1

4. a?

2a

+ 1

2x

-y

2/(2/-

-3a;)_

X

ar^.

-xy

a

+ &

a — b

28.
29.

a^ + 4a;-12 a:2-3a;-54

iK X

x'-Qax-^-^a? x^ + 4:ax-21a^

30 «' + ^' 9i k. 32 ^ 6 2 6^

"a--f-a6 a + 6 a a — 6 a + & o^ — b^

31. ^ + _^^ ^^. 33. -^ y- 1.

l + ic \—x 1— cc^ ic — 2/ aJ + y

34.
35.

FRACTIONS. 109

1 . 2x

a{a + x) a(a — x) a^ — x^

^ 2x 3x^ + 4

x + 2 {x + 2f (a; + 2f'

36 A 1 L 39 _J i2a~hr

a-3 a + Q a ' 2a + h Sa^ + ft^

- x-\-2 X ~2 _ 16 Ar. a-\- X _ a — x _ 4: ax
' x — 2 X + 2 x^ — 4 a~x a-j- x a? — x

38.

x + y o^ + f_ ^^ ^ 2(x + y) ^ {x + y)\

x — y 01? — if x — y ip^ — y)'

m —■ 71 (m — 7iy ' (m — w)^
x-\-l x~3 X — 5

42. — ^ v-^^^ +

43

x-\-2 X — 4: x^ — 2x

(a — b){b — c) (6 — c) (c — a) (c — a) (a — &)

cc — 3 a; — 2 a^— 5a; + 6
46. -^+ ^ ^«

a + & a — 6 a^ + 6^
^y _1 3 a; aa;

a — x a? — X? o? — X?

^Q ^; a , g^ — 4^

' a + 1 a^ — a + 1 a^ + l*

49.

a; + g ?/ + g a;4- y

(a? -y){y- 2) (^' - y) (^' -^) {^- ^) {y ~ ^)

50 a^ + 2 2(a;-l) a;-3

a? + 4x + 3 x' + x-Q, x''-x-2

In certain examples, the principles of §§ 130 and 131
enable us to change the form of a fraction so that the given
denominators shall be arranged in the same order of powers.

110 ALGEBRA.

51. Simplify -?-, + |^-
a — b b^ — a'

Changing the signs of the terms in the second denominator, at the
same time changing the sign before the fraction (§ 130), we have

3 2b + a

a-b a^-b^
The L. C. D. is now ct^ - b^.

Whence 3 2b + a _3(a + b)-(2b + a)

Whence, ^_^ ^^-ft^- a^ - 6'''

_Sa + Sb-2b-a_2a + b ^^^
a2 - 62 a-i - 62'

52. Simplify

(x-y)(x-z) {y-x){y-z) (z-x)(z-y)

By § 131, we change the sign of the factor y — x in the second de-
nominator, at the same time changing the sign before the fraction ;
and we change the signs of both factors of the third denominator.

The expression then becomes

1 .+ 1 1

(ic - ?/) (x - z) {x-y){y-z) {x-z){y-z)

The L. C. D. is now (x — y)(x — z^ {y — z) ; whence the result

_ (y-g) + (x-g) - (x-y) _ y-z-\-x-z-x+y
{x-y){x-z){y-z) ix-y)(x-s){y-s)

2y-2z _ 2(y-z)

{x-y)(x-z){y-z) (x-y){x-z)(y-z) (x-y){x-z)
Simplify the following:

63- -^ -T^- 57.

or — xy y' — xy

3x-6 8-4a;

55. ^ + -A_. 59.

56. — -i— -f— ^. 60.

, ^ns.

a

1

1

1

ab-b'

b — a

b

a

1

a

2

a + 1

'l

— a

a^-l

X

X

a^

2 + x

2

— X

0^-4

X

y

2^/2

4m — m^ m^ — 16 a^ + ^Z x — y y^ ~x^

FRACTIONS. Ill

61 1 _ 1 t 2a^ — 9 62 ^ *"■ ^ *^^

a 2 a — 3 9a — 4 a^ m-\-2 m — 2 4 —

63. 1^ + ^

(a — 6) (a + c) (& — a) (6 + c)
64. -^^ + ; ^"^ ^

65. , 1 , L + .

{^ -y){y- «) (2/ -x){x- z) {z -x){z- y)
66. ^ + ^ + . ^

(a — b)(a — c) (p — c)ih — a) (c — a) (c — 6)

MULTIPLICATION OF FRACTIONS.

143. Eequired the product of - and —

h d

Let ? X ^ = a;. (1)

b d

Multiplying eaeh of these equals hy bxd (%9, 1), we have

^ x-x b X d = X X b X d.
b d

Or, since the factors of a product may be written in any
order,

't xb\x(-xd]=x X b X d.

fi J \d

Whence, (a) x (c) =x xb x d. (§ 9, 3)

Dividing each of these equals by 6 x d (§ 9, 1), we have
axe

b X d

a .c axe

(2)

From (1) and (2). |x^ = ^. (§9,4)

We then have the following rule for the multiplication of
fractions :

Mrdtiply the numerators together for the numerator of the
product, and the denominators for its denominator.

112 ALGEBRA.

Common factors in the numerators and denominators
should be cancelled before performing the multiplication.

Integral or mixed expressions should be expressed in a
fractional form (§§ 135, 138), before applying the rule.

I/I/I 1 T\/r ^^-• ^ 10 d^V u^ 3 6V

144. 1. Multiply -^ by ^^•

We have 1^^ x ^^ = 2 x 5 x 3 x a^b*x^y ^ b^ ^^^^

' 9 6x2 4 a^yz 32 X 22 X a^hx'^y^ 6 y^
In this case, the factors cancelled are 2, 3, a^, b, x^, and y.

2. Find the product of -^ -, 2 - ^^, and x^ - 9.

ar + .x — 6 x — 3

We have, ? x (2 -5^") x (x2 - 9)

'x2 + cc-6Vx-3/

X ..2x-6-x + 8^(^2_9)

' x2 + X - 6 X

X

.x^±2x(x + 3)(x-3) = ^^^,^n5.

(x + 3)(x-2) x-3 ^ x-2

In this case, the factors cancelled are x + 3 and x — 3.

EXAMPLES.
Simplify the following :

14^ ^' 20n'x 2SxY ISm'^j

. &a?m 20 b^n* q c^+a-30 5 a

25 5 V Sa^m^ " 3 a a''-4.a-o

6. ^X:5ixi^. 10. ^^^xf+«^.

42/ 10 2; 9a; rrv^ — 25m3m — l

^ 4a« 156^ 21cf ,, a;2^3a._i8 2 0.-^-4 a-^

9&«^ 7c^ ^lOa^* ■ ar-8.tr+12 .x^ - 36

„ 3a263 6&V lOAi a;y + t/^ ar' + a;y-2y^^

4c^ 5a« 96^ .r^ _ ^^y ^ a^ + 2 a;^/ + y^

FRACTIONS. 113

a2.|-a6_20 5^^^ a^-4 6^ ^^ a;" - ^ ^^^ x" - 2a; + 4
a^— a6 — 6 6^ a^+5a6 ar^ + 8 cc^ + a^^ + o;

15. ^ . X — X

8a;-4 10a; + 5 3a;2_27
1 ft g^ + 2a o? — 16 a- + a

17.

3a — 4 a^ — a a'^^ + Ga + S

' 1^^^f\f^ 3a; + 5y
a;2_2/2 j^ 2a; + 32/.

18. (^ + y)^ ~ ^' X ^ ~ '^y ~ ^)^

(x — yf — z^ a^— {y-\- zf
19. ^-^ x^'~^'x^- ^^^

n/v a^a; + ax"^ a^ + 2 aa; + a^ g^ — 2 aa; + x^

a* — 2 c^x^ -\- x^ a^ -^ x^ ax

oi a;* — 1 4 a; + 3 A ,

21 . :rr— ; ^, X ^ „ ^ X ( 4 a; + ■

leaj^-g.^^ 2a^ + 2 V a;-l

DIVISION OF FRACTIONS.

145. Required the quotient of - divided by —

b d

Let ^^^ = x. (1)

Then since the dividend is the product of the divisor and
quotient (§ 54), we have

a c
0 a

Multiplying each of these equals by - (§ 9, 1), we have

-X- = -xxx- = x. (2)

bed c

From (1) and (2), ^ ^ ^ = ^ x ^- (§ 9, 4)

b d b c

114 ALGEBRA.

Therefore, to divide one fraction by another, multiply the
dividend by the divisor inverted.

Integral or mixed expressions should be expressed in a
fractional form (§§ 135, 138) before applying the rule.

146. 1. Divide 11^ by ^.

o afy* 10 x-y^

5xV lOx'^y'' 5xV Qffl'^fts Sb^

2. Divide 9 +-2^^ by 3 + -^.
ar — y^ *' x — y

We have,

Ix2-9y2 + 5y2 Sx-3y + by

i'-^H^^y

x^ — y^ x — y

_9x2j-4y2 ic-y _(3x + 2j/)(3x-2y) x-y

- x2 - j/2 ^ 3x + 2?/ ~ (X + »/) (x - y) ^ 3x + 2y

3x-2w ,
= ^, Ans.

x + y

EXAMPLES.
Simplify the following :

3_ 24a^6 : 8 a%l 7. /'-^-^Vf— + -

7ccY V^ft 2y V3& 2,

^ 21 m^^ . 14 gV g a" + 10 a + 21 . g^ - 9

10 b'm ' 15 b*m^' ' g^ _ 4 g^ + 3 a ' g^ - a^'

5. 3 . 2 g_ ^+4^H-4^_^^+2^

cc^— 6a;+8 a^— a;— 12 " ^~y ' x^ — xy

r, 4m^—25n^ . 2mn— 5n^ */v v? — x , cg^ — 2a; + l

16m^— 9n^ 4m^+3mn a^ + 1 ' cc^ — cc^ + a;

11.

. g^ + 2 g + 4

a^ + Ta + lO ■ a2 + 2g

g^ - 5 g6 - 14 &^ . g'' - 3 ab - 28 6^
a2 + 5g6-2462 " a^-8g6 + 156^

13.
14.

FRACTIONS.

(-

a'

a —
a —

■ 5x
■2x^

a?-

-b'

-0^-2 be

a

-b

— c

115

a2 _ ^2 _ g2 ^ 2 6c a + b

COMPLEX FRACTIONS.

147. A Complex Fraction is a fraction having one or
more fractions in either or both of its terms.

It is simply a case in division of fractions, its numerator
being the dividend, and its denominator the divisor.

148. 1. Simplify

b-^

We have, = -r-, = a x r-, (§ 146) = ^-5 , Ans.

h—- bd — c bd — c bd — c

d d

It is often advantageous to simplify a complex fraction
by multiplying its numerator and denominator by the lowest
common multiple of their denominators (§ 129).

2. Simplify \^ ^+^

+

a — b a -{-b
The L. C. M. of a + 6 and a-b is (a + b) (a- b).
Multiplying both terms by (a + 6)(a — &), we have

a(a + b)— a(a — b) _a^ + ab — a^ + ab _ 2ab j
b(a + b)+ a{a -b)~ ab + b'^+ d^ - ab~ a'^ + b-'

EXAMPLES.
Simplify the following :

a c -. . 1 1

n b d . 2m t-a^

3. 4. r— 5. T-.

a . c _ 1 ^1

-4-- m — - — 1

6 d 4m X

116 ALGEBRA.

m _ w ^ + 1 — ^ c»2 - 13 + —

g n m g y x_ j^^ ^^

m?i y X X

^_2-)-^ <^ g — & a? a; — ?/

2 3' " 6 g — 6 ' aj x + y

y X a a + h ^ — y ^

a-x a? Sy 2b'

a — -J — ^ a -

g 1 + ax jj 3/ a; ^^ ft-o

l + ax* y X a + 26

2 , , 1 , 2 2a + &^2a-&

a — 1 a + od> a — 6b

15. q • lb. ;; — ^ n T"

1 2a-\-b 2a— b

a — 1 a — 36a + 36

1

17. Simplify .

1+ —

We have, -^ = _^ = _^_ = ^-^^ , ^ns.

1+1 ^+1

In examples like the above, begin by simplifying the lowest complex

1 X

fraction; first multiply both terms of by x, giving ^ , and

1 + -

1 "^ X + \

then multiply both terms of by x + 1, giving

j.x ■ x+l+x

X + 1

Simplify the following :

18. 3 ^-^. 19. 1-

5+ ^

7+^

X 1 — a

FRACTIONS. 117

^ Sjd' + b^ x + y a^ + y'

20. ^^'-^' 23 •''""^ "^-y'

^ 2(a + 2&)' ■ _^_a^^^'_+^'

3 a + 6 a; + 2/ (a; + i/)^

a; + ax — a 2n(m — w)

x — a x + a 2^ ?» +^

a^ + a^ _ ^ ' ■wi^ + 7i^

(a; — of mn + w^

1 - a.-^ 1+a? ft 4- & gs 4- 63

1 + ar^ l_a^ a - & a^ - 6«

1-a; 1 + a;' aj-^ oN-^!'

l + a;l — a; a — ha? — h^

MISCELLANEOUS AND REVIEW EXAMPLES.
149. Reduce each of the following to a fractional form :

1. 9a-2-l!i^^'. 2. -^±^^(c, + ai + V)
4 a + 3 o?-ah + h^

Simplify the following :

n 1 2a 6aa;

2a-3a; (2a-3a:)2 (2a-3a;)

(l-a;)(l-y) ^ (a^-2)^-a^

(1 + a:?/)' - (a: + 2/)' ' a*-3a2-4

V32/' a:?/ 4ary ' \3y 2x

Q a-\-b , c + d . 2 (ac — 6c^)
a — 6 c — d (b — a)(c — d)

Qxy-(x + 2yy jq (a^ _ 6 a; - 4)^ - 144

a^ + 8/ ■ ■ (x^ -h a; - 11)2 - 81 '

V

118

11.

ALGEBRA.

3a

4a

a2-3a + 2"^a2-7a + 10 a^-6a + 5

13.
14.

12. /2a + 3--2i^\
\ 2a + 3/

x^ + y^ x^ — if

a;2 _ 2/2 £c3 _|. 2/3

2a-3 +
15

24 a

2a-3

a* + a^6 — a^&^ — ab*
a*b - a^b' + ab* - b''

17.
18.
19.

]f y X ^J\y X

J/J o?cd + a&(^^ — ah(? — fe-cd
a'^cd — abd? — a&c^ + 6^cd

;-3 ^

3a; -8 a; + 2

\» — 2 a; —

Z' 2 i_\r 1 i_\

^^a^ 4.3a; + 2 x-^\) \x'-^^x + 2 x + 2j

m

m

2 m*

2(m — n) 2(m + n) n^(m^ — n^

20 (o^ + ^ + c)' - (« + & - c)^
(a-b + cy-(a-b-cy

'\\j 2a(2a-36a;)+35(35 4-2aa;)
(2a-3&a;)2+(36 + 2ax)2

21.
22

a — 6 —

a; — y , a;4- z _ (y + ^)^

a;_2/ (x-y){x + z)

x-{-z

a?b — a&^

{a + bf
*''• a + 6 a'' + 6'' '

24.

a — 6
25. -

a^-ft^
1 .

4a^-16a^ + 17a;-3
6a;3_i7a32 + 8a; + 6*

a

— V

a; + 2 ay a^ + ax — 2 a''

26.
27.

a2 _ 4a6 + 46^ ' (^a^ - 46^ a" + ab - 667*

X* + 2 x^y^ -hy* \ xy

FRACTIONS

119

28.

a; — 1 x + 1

- +. 1

1 OJ + l

{x^ + 2x-l).

nq m^ + 2 mn -f 4 rt^ 4 m^ — 9 w^ . 2 ?/i + 3 n

30.

2m — 3n
1

7?i^ — 8 n^ m" — 4 n^

a; + 2/ ix^ — y^

a{a - b)(a - c) ^ &(6 - c)(6 - a)
i-32. ^+^ ^+_^

+

. 31 "'^

-y

3.3 _|_ y^

'^^' x^

+ f

x-y

x"

y +

z

x + y

(«-y)(2/-2) 0(-2)(2-a') {z~x){x-y)

33.

\ a?-ab + h-]\

ah

a'' + 2ab+by a^ -

a^ + b^

34.

' +^^J-.+ '

X 1+x 1+x^ l-{-x*

(First add the first two fractions ; to the result add the third frac-
tion, and to this result add the fourth fraction.)

35.
36.

a , a , 2 a? ^ 4 a*

a -2 a + 2 0^ + 4. a* + 16

_i L_+_J L_.

x — l cc + l a; — 2 a; + 2

(First combine the first two fractions, then the last two, and then
add these results. )

37.
38.
39.

1 2a

2a

a—h a + b ' a^ — b"^ a^ + b-

_JL 1 3a^ Sx"

x-1 a; + la^ + l ar'-l'

4a;-3

: + :

2a;-5

6x2 + 13a;-5 12a^ + 5a;-3
(Find the L. C. M. of the denominators by the method of § 126.)
3a4-2 5a-l

40.

6a'' -a- 12 lOa^-lOa + G

120 ALGEBRA.

XIII. SIMPLE EQUATIONS (Continued).

SOLUTION OF EQUATIONS CONTAINING FRACTIONS.

150. Clearing of Fractions.

Consider the equation =

^ 3 4 6 8

The lowest common multiple of 3, 4, 6, and 8 is 24.
Multiplying each term of the equation by 24 (§ 71, 2),

we have

16x-30 = 20a;-27,

where the denominators have been removed.

We derive from the above the following rule for clearing
an equation of fractions :

Multiply each term by the lowest common multiple of the
given denominators.

151. 1. Solve the equation -^ — - = — — -.

6 3 5 4

The L. C. M. of 6, 3, 5, and 4 is 60.

Multiplying each term of the equation by 60, we have

70X-100 =36x-15.
Transposing, 70 x - 36 x = 100 - 15.

Uniting terms, 34x = 85.

Dividing by 34, x = — = -, Ans.

o J > 34 2

EXAMPLES.
Solve the following equations :

2. x+l-^4 = ^. 3. ^^-^ + ^ = 0.

2 5 d 4 6

SIMPLE EQUATIONS. 121

Sx

2 ■

X 5a; 1
3~ 4 ^8'

4:X

9

2 5a; 3a;

3 6 2

7x
2

4a; 2a;_ 11
3 5 6

_3_
5x

-l = l-±.

10 4 a;

5

1

_1
4

8 _
'9 a;"

18 a;

6a;

3.

2 ^

5
14

X _
3~

7x
6

4a;

Y'

^ f\ ^ OC a!/|0 lit/ 0«t/

y~2 lo^'s r*

11 A_A_A + A = J_.

3a; 4a; 5a; 6a; 20

If a fraction whose numerator is a polynomial is preceded
by a — sign, it is convenient, on clearing of fractions, to
enclose the numerator in a parenthesis, as shown in Ex. 12.

If this is not done, care must be taken to change the
sign of each term of the numerator when the denominator is
removed.

12. Solve the equation — — — = 4 H -~ —

4 5 10

The L. C. M. of 4, 5, and 10 is 20 ; multiplying each term by 20,

we have

15x-6-(16x-20) = 80 + Ux + 10.

Whence, 15x - 5 - 16x + 20 = 80 + 14x + 10.

Transposing, 15 x - 16 x - 14 x = 80 + 10 + 5 - 20.

Uniting terms, — 15 x - 75.

Dividing by — 15, x = — 5, Ans.

Solve the following equations :

13. 40; + ^^-^^ = ^. 16.

7 ^

14. ^_2^_2 = ^_2. 17.

3 9

15. 2a;-^^±^ = |+l. 18.

19 lla; + 4 14x + 3 10a; + 7_Q
2 4 8 ■

3

X

a; + 7
3

8a;-4 ^

7

2a;-

5 3;

x-8 2

7

6 3

a; + 2

9

3a; + 14

10

35

14

122 ALGEBRA.

2 8^^ 12

5^ ^ 20 5 2

oo 10(a; + 2) 5a; -4 5a; + 12 _.,i
9 12 6 ~ ''

2 3^ ^6^ ^ 4

24. ^^ ^ - ^ - 1 (3 a; - 1) = ^^ ^' + '^ - ^ (7 a; - 2).

3 2^ • ^ 6 9^ ^

25.
26.

2a; + 4 7x-1^13a; + 5 lla;~3_
5 2 3 10 *

7 a; — 8 _ 7a; + 6 ^ a? — 5 _ 4a; + 9.
14 4a; ~ 2 7a; '

27 3(a;-3) 2(a;^-5) 5a;'^-12^ 9
, ' 2 3a; 6x' 2

2 5 2
28. Solve the equation = 0.

^ a;-2 a; + 2 a:^ — 4:

The L. C. M. of X - 2, X + 2, and x- - 4 is x^ - 4.
Multiplying each term by x^ — 4, we have
2(x + 2)-5(x-2)-2=0.
Whence, 2x + 4 - 5x + 10 - 2 = 0.

Transposing, 2x — 5x= — 4 — 10 + 2.

Uniting terms, — 3x = — 12.

Dividing by — 3, x = 4, Ans.

If the denominators are partly monomial and partly poly,
nomial, it is often advantageous to clear of fractions at first
partially; multiplying each term of the equation by the
L. C. M. of the monomial denominators.

29. Solve the equation

SIMPLE EQUATIONS. 123

6a;4-l 2a;-4 2a;-l

15 7 a; - 16

The L.C. M. of the monomial denominators is 15.
Multiplying each term by 15, we have

6 X + 1 = o X — o.

7x- 16

Transposing and uniting terms, 4 = — — — ^'

Multiplying by 7 x - 16, 28 x - 64 = 30 x - 60.
Transposing, 28 x - 30 x = 64 - 60.

Uniting terms, — 2 x = 4.

Dividing by — 2, x= —2, Ans.

Solve the following equations :

30 — ^ ^ = 0. 35. L^±10^=24__5_^7

5a; + 2 3a; + 4 (x + lf x+1

2^ 2a; + 3^4a; + 5, 33 6.t+7^^ ^ 2

3a;-4 6a;-l 6 2(2a;+l)

32 15 g-^- 5a; -8 ^ ^ 37 ^ + ^^ + '^ = 9a;-2

3a.-2 + 6a; + 4 " v^ ' 3 3a;-4 9

33. 6^ + 5_^3^^ W^8. -5 2_^_J_.

2a;(a;-l) a^-l'Y^ 3a;-5 a;-2 a;-3

34 3a; 2a; _2a;-— 5 gg 2a; + 7 5a; — 4_a; + 6

2a;+3 2a;-3~'4ar'-9" ' 14 3a; + l~ 7

40.

3 4

a;-2 2a;-l 3a; + 2

41 2(a;-7) a; - 2 cc + S^^
* ar^ + 3a;- 28 a;-4 a; + 7

42 5a; + l 4 a; + 7 3-^ — 2_a

5 6a; + ll 3

43. ,r^+ ^ ^

2a; + 3 3a;-2 4a; + l

124 ALGEBRA.

44.

2cc + l 2a.--l 9a7 + 17

2a; -16 2rc + 12 x2-2a;-48

45. ?^_^^^2jM^^^
a; + 2 a; + 1 .^'2 - 1

2 + 3a; 2-3.r 36-4a;

46.

3 - a; 3 + a; a;^ - 9

48.

,- 2ar' + 3.'K-l , 2.a;S-3a; + l ^ ,

(a;+l)(a; + 3)^a;-6 7(3a;-8) 3(a;-2)^

(a; + 5)(a; + T) ^ + 2* 3(a.- - 3) "^ 3.r - 1

^Q 3a^+5a;-4^3.^•+5 -^ 2a;+7 3a;-5^17a;+2
4ar^-3a;4-2 4a;-3' ' 6a;-4 9a;+6 9ar*-4*

a; — 2 » — 3 a; — 5 x — 6

52.

3 a; — 4 a;— 6 a; — 7

(First combine the fractions in tiie first member ; then the fractions
in the second member.)

7a; + 3
14 28 4(4 a; -3)*

SOLUTION OF LITERAL EQUATIONS.

152. A Literal Equation is one in which some or all of
the known quantities are represented by letters ; as,

2a; + a = 6ar'-10.

153. 1. Solve the equation {b — cxf—{a — cxf= h(h — a).
Performing the operations indicated, we have

62 - 2 6cx + c2x2 - (a2 _ 2 acx + chfl) = b- - ah.
Whence, lfi-2hcx-\- dhi^ -a^+2acx- c^x-^ = b'^ - ab.
Transposing and uniting terms, 2acx — 2 6cx = a" — ab.
Factoring both members, 2cx(a — b)= a(a — b).

Dividing by 2 c(a - &), x = "^" ~ ^^ = — , Ans.

SIMPLE EQUATIONS. 125

EXAMPLES.
Solve the following equations :

2. a(3bx-2a)=h(^2a-3bx).

3. (a; + ay + (?> + cf = (x - «)2+ (b - cf.

. X — a , 2x o n x^ — b b —X 2 x b

x X— a ax a a X

5 3a;-4^5m -2n ^ ^ ^ .t - 2 ^ 1 _

" 3 £c + 4 5 wi, + 2 ?i ' m w?

n 5x — 2a 9x — 5a^ 3 (a; + 2 a'^) 5a; _ ^
2 a 3a^ a^ 6 a

q ax — b bx-\-a__c) a — 6_

10. 2(a; - 6) (2 a - 3 6 - 3 ;r)-(2 a - 3 a;) (6 + 2 a;)= 0

11. (a; + m) (x + n) — (x ~ m) (a; — ?i) = 2 (m + 7i)^.

12 a; — & a; 4- ^ _ 4 <t' — &^

a; — 2a a; + 2a a;^— 4a^

j3 2a; + 3a_3a; + 46 j^ 2 ?ia-' — 3 ^ ^ 9 nx + 2

2a; — 3& 3a; — 4a wa; — 1 37ia; — 1

tc a — b , b — c , c — a ^
^, ID. 1 1 = U.

it/ C *t/ Cfc U/

16. (a; + a/ + (a;-a)='=2a;(a;2_^2>)_24£v3.

, . , <""
, „ 3 a;(a — 6) _ a — 26 a — 6 _ n ~
a? — b"^ x + b b — X

, o a; a — 2 bcx _ 5 a; _ 8 ac — 8 bx — 9 a
2 4 &c "" e'c i2&^

19. (a; - 2 a + 3 6)- - (a; - 2 a) (x + 3 6) - 6 a^; := 0.

126 ALGEBRA.

2Q __a &_^ _ W — a? ^ nn (2 a; — 3 mf _ x — 3m

x — a X — b b'^ — bx (2 a? — 3 n)'^ x — 3n

21. -^^ + J^ = a + b. 23. ^^^^+^^) = f!i±^--:?i:z^.
a; + 6 x + a x x—b x+a

24. _!-+ ^ 2.^•-a-&

a; — a x — b xix — a — b)

t; + 4 g + 5 4 a- + a + 2 i
a; + a + & x -\- a — b

OK .^' + 4 g + 5 4 a; + a + 2 ?> _ ^^

SOLUTION OF EQUATIONS INVOLVING DECIMALS.
154. 1. Solve the equation .17 a; - .23 = .113 a; + .112.

Transposing, .17 x - .113x = .23 + .112.

Uniting terms, .057 x = .342.

Dividing by .057, x = ~ = 6, Ans.

.057

EXAMPLES.
Solve the following equations :

2. 2.9 a;- 1.98 = 1.4 a;- 1.845.

3. .05 a; + .117 = .186 a; -.2 a; -.139.

4. .6 a; -.265 + .03 = .4 + .66 a- -.187 a;.

5. .4(L7.a;-.6) =.95a; + 5.16.

6. .08(35 a; -2.3) = .9(7 a; + .18) -.997.

« OQ .29 a; +.0184 ^ ^^,

7. 2.8 a; ' = .5 a; — .064.

8.

ggg .4a; + .708_18 .3
2 a; 5 a;

9.

.7 a; + .371 .3 a; -.256 ,^
.9 .6 ■ ■

10.

2-3a; 3a;-14 a;-2 10.t

-9

1.5 9 1.8 2.25

SIMPLE EQUATIONS. 127

PROBLEMS.

155. 1. Divide 43 into two parts such that three-eighths
of one part may equal two-ninths of the other.

Let z = one part.

Then, 43 — x = the other.

3x ''
By the conditions, — = - (43 — x).

Clearing of fractions, 27 x = 16 x 43 — 16 x.

Transposing, 43x = 16 x 43.

Dividing by 43, x = 16, one part.

Whence, 43 — x = 27, the other part.

2. The fifth part of a number exceeds its eighth part by
3 ; what is the number ?

3. What number is that from which if four-sevenths of
itself be subtracted, the result will equal three-fourths of
the number diminished by 18 ?

4. What number exceeds the sum of its third, sixth, and
fourteenth parts by 18 ?

5. Divide 45 into two parts such that the sum of four-
ninths the greater and two-thirds the less shall equal 24.

6. Divide 56 into two parts such that five-eighths the
greater shall exceed seven-twelfths the less by 6.

7. Divide f 124 between A, B, and C so that A's share
may be five-sixths of B's, and C's nine-tenths of A's.

8. A man travelled 768 miles. He Avent four-fifths as
many miles by water as by rail, and five-twelfths as many
by carriage as by water. How many miles did he travel in
each manner ?

9. A's age is three-eighths of B's, and eight years ago
it was two-sevenths of B's age ; find their ages at present.

128 ALGEBRA.

10. A has $ 52, and B $ 38. After giving B a certain
sum, A has only three-sevenths as much money as B. What
siun was given to B ?

11. I paid a certain sum for a picture, and the same price
for a frame. If the picture had cost $ 4 more, and the
frame 30 cents less, the price of the frame would have been
one-third that of the picture. Find the cost of the picture.

12. A can do a piece of work in 8 days which B can per-
form in 10 days. In how many days can it be done by both
working together ?

Let X — the number of days required.

Then, - = the part both can do in one day.

Also, - = the part A can do in one day,

8

and — = the part B can do in one day.

By the conditions, - H = -•

8 10 X

5x + 4x = 40.

9x=:40.

Whence, x = 4|, the number of days required.

13. The second digit of a number exceeds the first by 2 ;
and if the number, increased by 6, be divided by the sum of
its digits, the quotient is 5. Find the number.

Let X — the first digit.

Then, x + 2 = the second digit,

and 2 X -h 2 = the sum of the digits.

The number itself is equal to 10 times the first digit, plus the second
Then, lOx 4-(a; + 2), or 11 x + 2 = the number.

By the conditions, ^^^ + ^ + ^ = 5.
2x + 2

llx + 8 = lOx-1- 10.

Whence, x = 2.

Then, 11 x -)- 2 = 24, the number required.

SIMPLE EQUATIONS. 129

14. A can do a piece of work in 18 days, and B can do
the same in 2-1 days. In how many days can it be done by
both working together ?

15. A can do a piece of work in 3.V hours which B can do
in 3| hours, and C in 3| hours. In how many hours can
it be done by all working together ?

16. A tank can be filled by one pipe in 9 hours, and
emptied by another in 21 hours. In what time will the
tank be filled if both pipes be opened ?

17. A vessel can be filled by three taps;, by the first
alone in 7^ minutes, by the second alone in 4i minutes, and
by the third alone in 4| minutes. In what time will it be
filled if all the taps be opened ?

18. The first digit of a number is 4 less than the second ;
and if the number be divided by the sum of its digits, the
quotient is 4. Find the number.

19. The second digit of a number is one-fourth of the
first; and if the number, diminished by 10, be divided by
the difference of its digits, the quotient is 12. Find the
number.

20. If a certain number be diminished by 23, one-fourth
of the result is as much below 37 as the number itself is
above 56. Find the number.

21. What number is that, seven-eighths of which is as
much below 21 as three-tenths of it exceeds 2i ?

22. B is 24 years older than A ; and when A is twice his
present age, B will be | as old as he now is. How old is
each?

23. The denominator of a fraction exceeds the numerator
by 5. If the denominator be decreased by 20, the resulting
fraction, increased by 1, is equal to twice the original frac-
tion. Find the fraction.

130 ALGEBRA.

24. Divide 44 into two parts such that one divided by
the other shall give 2 as a quotient and 5 as a remainder.

Let X = the divisor.

Then, 44 — a; = the dividend.

Now since the dividend is equal to the product of the divisor and
quotient, plus the remainder, we have

44 - a; = 2 X + 5.

-3x = -39.

Whence, x = 13, the divisor,

and 44 — X = 31, tlie dividend.

25. Two persons, A and B, 63 miles apart, start at the
same time and travel towards each other, A travels at the
rate of 4 miles an hour, and B at the rate of 3 miles an
hour. How far will each have travelled when they meet ?

Let 4x = the number of miles that A travels.

Then, 3x = the number of miles that B travels.

By the conditions,

4x + 3x = 63.

7x = 63.

x = 9.

Whence, 4x = 36, the number of miles that A travels,

and 3 X = 27, the number of miles that B travels.

Note, It is often advantageous, as in Ex. 25, to represent the
unknown quantity by some multiple of x instead of by x itself.

26. Divide 49 into two parts such that one divided by
the other may give 2 as a quotient and 7 as a remainder.

27. Two men, A and B, 66 miles apart, set out at the
same time and travel towards each other. A travels at the
rate of 15 miles in 4 hours, and B at the rate of 9 miles
in 2 hours. How far will each have travelled when they
meet ?

SIMPLE EQUATIONS. 131

28. Divide 134 into two parts such that one divided by
the other may give 3 as a quotient and 26 as a remainder.

29. The denominator of a fraction is 7 less than the
numerator ; and if 5 be added to the numerator, the value
of the fraction is f. Find the fraction.

30. The second digit of a number exceeds the first by 4 ;
and if the number, increased by 39, be divided by the sum
of its digits, the quotient is 7. Find the number.

31. I paid a certain sum for a horse, and seven-tenths as
much for a carriage. If the horse had cost $ 70 less, and
the carriage $ 50 more, the price of the horse would have
been four-fifths that of the carriage. What was the cost of
each ?

32. A can do a piece of work in 15 hours, which B can
do in 25 hours. After A has worked for a certain time, B
completes the job, working 9 hours longer than A. How
many hours did A work ?

33. A man owns a horse, a carriage worth $100 more
than the horse, and a harness. The horse and harness are
together worth three-fourths the value of the carriage, and
the carriage and harness are together worth $ 50 less than
twice the value of the horse. Find the value of each.

^ 34. The rate of an express train is f that of a slow train,
and it covers 180 miles in two hours less time than the
slow train. Find the rate of each train.

35. Two men, A and B, 57 miles apart, set out, B 20
minutes after A, and travel towards each other. A travels
at the rate of 6 miles an hour, and B at the rate of 5
miles an hour. How far will each have travelled when
they meet ?

36. A grocer buys eggs at the rate of 4 for 7 cents. He
sells one-fourth of them at the rate of 5 for 12 cents, and
the remainder at the rate of 6 for 11 cents, and makes 27
cents by the transaction. How many eggs did he buy ?

132 ALGEBRA.

37. At what time between 3 and 4 o'clock are the hands
of a watch opposite to each other ?

Let X = the number of minute-spaces passed over by the minute-
hand from 3 o'clock to the required time.

Tlien, since the hour-hand is 15 minute-spaces in advance of the
minute-hand at 3 o'clock, x — 15 - 30, or x - 45, will represent the
number of minute-spaces passed over by the hour-hand.

But the minute-hand moves 12 times as fast as the hour-hand.

Whence, x = 12 (x — 45).

X = 12 X - 540.

- llx = -540.

X = 40 J-.
Then the required time is 49^^ minutes after 3 o'clock,

38. At what time between 1 and 2 are the hands of a
watch opposite to each other ?

39. At what time between 6 and 7 is the minute-hand of
a watch 5 minutes in advance of the hour-hand ?

40. At what time between 4 and 5 are the hands of a
watch together ?

41. At what time between 5 and 5.30 are the hands of a
watch at right angles to each other ?

42. The sum of the digits of a number is 15 ; and if the
number be divided by its second digit, the quotient is 12,
and the remainder 3. Find the number.

43. A man has 11 hours at his disposal. How far can
he ride in a coach which travels 4i miles an hour, so as to
return in time, walking back at the rate of 3f miles an hour ?

44. A, B, and C together can do a piece of work in 1|
days ; B's work is one-half of A's, and C's three-fourths of
B's. How many days will it take each working alone ?

45. At what time between 9 and 10 are the hands of a
watch together ?

SIMPLE EQUATIONS. 133

•

46. A, B, C, and D found a sum of money. They agreed
that A should receive $4 less than one-third, B f 2 more
than one-fourth, C $3 more than one-fifth, and D the
remainder, $ 25. How much did A, B, and C receive ?

•47. At what time between 8 and 9 are the hands of a
watch opposite to each other ?

48. A vessel can be emptied by three taps ; by the first
alone in 90 minutes, by the second alone in 144 minutes,
and by the third alone in 4 hours. In what time will it be
emptied if all the taps be opened ?

49. A and B start in business, B putting in f as much
capital as A. The first year, A loses f 500, and B gains \
of his money; the second year, A gains \ of his money,
and B loses $ 205 ; and they have now equal amounts.
How much had each at first ?

50. A man buys two pieces of cloth, one of which con-
tains 6 yards more than the other. For the larger he pays
at the rate of f 7 for 10 yards, and for the smaller at the
rate of f 5 for 3 yards. He sells the whole at the rate of
9 yards for $ 11, and makes $ 5 on the transaction. How
many yards were there in each piece ?

51. A man loaned a certain sum for 3 years at 5 per cent
compound interest; that is, at the end of each year there
was added -^ to the sum due. At the end of the third
year, there was due him $ 2130.03. Find the amount
loaned.

52. At what times between 7 and 8 are the hands of a
watch at right angles to each other ?

53. At what time between 2 and 3 is the hour-hand of a
watch one minute in advance of the minute-hand ?

54. Gold is 19^ times as heavy as water, and silvt^r 10^
times. A mixed mass weighs 19(i0 oz., and displaces 120 oz.
of water. How many ounces of each metal does it contain ?

134 ALGEBRA.

55. A merchant increases his capital annually by one-
third of it, and at the end of each year takes out $ 1800
for expenses. At the end of three years, after taking out
his expenses, he finds that his capital is $ 3800. V/hat was

his capital at first ?
/
"^56. A and B together can do a piece of work in 2| days,

B and C in 2^ days, and C and A in 2|- days. How many

days will it take each working alone ?

57. A alone can do a piece of work in 15 hours ; A and
B together can do it in 9 hours, and A and C together in 10
hours, 'a commences work at 6 a.m. ; at what hour can he
be relieved by B and C, so that the work may be completed
at 8 P.M. ?

58. A man invests -j^ of a certain sum in 4|- per cent
bonds, and the balance in 3^ per cent bonds, and finds his
annual income to be $ 117.50. How much does he invest
in each kind of bond ?

( The annual income from p dollars, invested at r per cent, is rep-
resented by ^^. 1
100 ^

59. An express train whose rate is 36 miles an hour starts
54 minutes after a slow train, and overtakes it in 1 hour
48 minutes. What is the rate of the slow train ?

60. At what time between 10 and 11 is the minute-hand
cf a watch 25 minutes in advance of the hour-hand ?

61. A woman sells half an egg more than half her eggs.
She then sells half an egg more than half her remaining
eggs. A third time she does the same, and now she has
sold all her eggs. How many had she at first ?

62. A man invests two-fifths of his money in 6^ per cent
bonds, two-ninths in 5\ per cent bonds, and the balance in
3|- per cent bonds. His income from the investments is
$ 915. Find the amount of his property.

SIMPLE EQUATIONS. 135

63. A man starts in business with f 8000, and adds to
his capital annually one-fourth of it. At the end of each
year he sets aside a fixed sum for expenses. At the end of
three years, after deducting the fixed sum for expenses,
his capital is reduced to $ 6475. What are his annual
expenses ?

64. If 19 oz. of gold weigli 18 oz. in water, and 10 oz. of
silver weigh 9 oz. in water, how many ounces of each metal
are there in a mixed mass weighing 127 oz. in air, and
117 oz. in water ?

65. A fox is pursued by a hound, and has a start of 63
of her own leaps. The fox makes 4 leaps while the hound
makes 3 ; but the hound in 5 leaps goes as far as the fox
in 9. How many leaps does each make before the hound
catches the fox ?

(Let 4x = the number of leaps made by the fox, and 3x = the
number made by the hound.)

66. A merchant increases his capital annually by one-
third of it, and at the end of each year sets aside f 2700 for
expenses. At the end of three years, after deducting the
sum for expenses, he has ^ of his original capital. Find
his original capital.

PROBLEMS INVOLVING LITERAL EQUATIONS.

156. 1. Divide o into two parts such that m times the
first shall exceed n times the second by b.

Let X = one part.

Then, a ~x = the other part.

By the conditions, mx = n(a — x) + 6.

mx = an — nx + b.

mx + nx = an + b.

x(m + n) = an -h b.

136

Whence,
and (

ALGEBRA.

X

_an + b
m + n'

one part,

an

+ h
+ n

_ am + an — an — b

m

in

+ n

am — b

— '

the other

part.

m + n

Note. The results can be used as formulas, for solving any prob-
lem of the above form. ^

Thus, let it be required to divide 25 into two parts such that 4 times
the first shall exceed 3 times the second by 37.

Here, a — 25, m = 4, n = 3, and 6 = 37.

Substituting these values in the results of Ex. 1 ,

the first part = ^A^^A±IL = 75 + 37 ^U2 = i6,

7 7 7

, ,, , ^ 25 X 4 - 37 100-37 63 ^
and the second part = = = — = 9.

7 7 7

2. Divide a into two parts such that m times the first
shall equal n times the second.

3. A is m times as old as B, and a years ago he was
n times as old. Find their ages at present.

4. A can do a piece of work in m hours, which B can
do in n hours. In how many hours can it be done by both
working together ?

5. A vessel can be filled by three taps ; by the first
alone in a minutes, by the second alone in b minutes, and-
by the third alone in c minutes. In how many minutes
will it be filled if all the taps be opened ?

6. A has m dollars, and B has n dollars. After giving
A a certain sum, B has r times as much money as A. What
sum was given to A ?

7. A gentleman distributing some money among beggars,
found that in order to give them a cents each, he would need
b cents more. He therefore gave them c cents each, and had
d cents left. How many beggars were there ?

SIMPLE EQUATIONS. 137

8. A man has a liours at his disposah How far can he
ride in a coach which travels b miles an hour, so as to return
home in time, walking back at the rate of c miles an hour ?

9. A courier who travels a miles in a day is followed
after n days by another who travels b miles in a day. In
how many days will the second overtake the first ?

10. What principal at r per cent interest will amount to
a dollars in t years ?

11. In how many years willp dollars amount to a dollars
at r per cent interest ?

12. At what rate per cent will p dollars amount to a dol-
lars in t years ?

13. Divide a into two parts, such that one divided by the
other may give & as a quotient and c as a remainder.

14. Two men, A and B, a miles apart, start at the same
time, and travel towards each other. A travels at the rate
of m miles an hour, and B at the rate of w miles an hour.
How far will each have travelled when they meet ?

15. A grocer mixes a pounds of coffee worth m cents a
pound, b pounds worth n cents a pound, and c pounds worth
p cents a pound. Find the cost per pound of the mixture.

\'"16. A banker has two kinds of money. It takes a pieces
of the first kind tc make a dollar, and b pieces of the second
kind. If he is offered a dollar for c pieces, how many of
each kind must he give ?

17. Divide a into three parts, such that the first may b£
m times the second, and the second n times the third.

18. A and B together can do a piece of work in m hours,
B and C in 7i hours, and C and A in j9 hours. In how many
hours can each alone do the work ?

X38 ALGEBRA.

XIV. SIMULTANEOUS EQUATIONS.

CONTAINING TWO UNKNOWN QUANTITIES.

157. If a rational and integral monomial (§ 69) involves
two or more letters, its degree with respect to them, is denoted
by the sum of their exponents.

Thus, 2a^b^xy^ is of the fourth degree with respect to
X and y.

158. If each term of an equation containing one or more
unknown quantities is rational and integral, the degree of
the equation is the degree of its term of highest degree.

Thus, if X and y represent unknown quantities,

ax — hy = ciB an equation of the first degree.

a? -\- A: X = — 2 is an equation of the second degree.

2 a^ — 3 xy"^ = 5 is an equation of the third degree ; for the
term 3 xy^ is the term of highest degree, and 3 xy^ is of the
third degree.

159. An equation containing two or more unknown quan-
tities is satisfied by an indefinitely great number of sets cf
values of these quantities.

Consider, for example, the equation x -\-y =-5.

If ic = 1, we have 1 -f 2/ = 5, or 2/ = 4.

If 05 = 2, we have 2 -\-y = b, ox y = ^\ and so on.

Thus the equation is satisfied by any one of the sets of

values

x = \, ^ = 4;

x=2, y = 3 ; etc.

Eor this reason, an equation containing two or more un*
known quantities is called an indeterminate equation.

SIMULTANEOUS EQUATIONS. 139

160. Two ec^uations, each containing two unknown quan-
tities, are said to be Independent when one of them is satis-
fied by sets of vakies of the unknown quantities which do
not satisfy the other.

Consider, for example, the equations x + y = 5, x — y = 3.

The first equation is satisfied by the set of values
x = 3, y = 2, which does not satisfy the second.

Therefore, the equations are independent.

But the equations x -\-y = 5, 2 x -\-2y = 10, are not inde-
pendent; for the second equation can be reduced to tlie
form of the first by dividing each term by 2 ; and hence
every set of values of x and y which satisfies one equation
will also satisfy the other.

161. Let there be two independent equations (§ 160),
each of the first degree, containing the unknown quantities
X and y, a.s x -\- y = 5, x — y = 3.

By § 159, each equation considered by itself is satisfied
by an indefinitely great number of sets of values of x and y.

But there is only one set of values of x and y, i.e., x = 4.,
y = 1, which satisfies both equations at the same time.

A series of equations is called Simultajieous when each
contains two or more unknown quantities, and every equa-
tion of the series is satisfied by the same set of values of
the unknown quantities.

162. To solve a series of simultaneous equations is to find
the set of values of the unknown quantities involved which
satisfies all the equations at the same time.

163. Two independent, simultaneous equations may be
solved by combining them in such a way as to form a single
equation containing \)ut one unknown quantity.

This operation is called Elimination^___

There are three principal metho33 of elimination.

5x-Sy = 19.

(1)

7x-\-4:y= 2.

(2)

20x - 12 ij= 76.

(3)

21x+12?/= 6.

(4)

140 • ALGEBRA.

164. I. Elimination by Addition or Subtraction.

1. Solve the equations

Multiplying (1) by 4,
Multiplying (2) by 3,

Adding (3) and (4), 41 x = 82.

Whence, x = 2.

Substituting the value of x in (1), 10 — 3y = 19.

- 3 .V = 9.
Whence, ?/ = — 3.

The above is an example of elimination by addition.

2. Solve the equations

Multiplying (1) by 2,
Multiplying (2) by 3,

Subtracting (4) from (3),
Whence,

Substituting the value of y in (2), 10 x — 14 = — 24.

10x=-10.
Whence, x =— 1.

The above is an example of elimination by subtraction.

Rule.

If necessary, multiply the given equations by such numbers
as will make the coefficients of one of the unknown quantities
in the resvUing equations of equal absolute value.

Add or subtract the resulting equations according as the
coefficients of equal absolute value are of u) dike or like sign.

Note. If the coefficients which are to be made of equal absolute
value are prime to each other, each may be used as the multiplier for
the other equation ; but if they are not prime to each other, such
multipliers should be used as will produce their lowest common mul-
tiple. Thus, in Ex. 1, to make the coefficients of y of equal absolute
value, we multiplied (1) by 4 and (2) by 3 ; but in Ex. 2, to make the
coefficients of .^ of equal absolute vali>e, since the L. C. M. of 10 and
16 is 30, we multiplied (1) by 2 and (2) by 3.

'15x + 8y =

\l0x~7y = -

30x+lQy= 2.
30x-21?/=-72.

1.

-24.

(1)
(2)

(3)
(4)

37 2/ = 74.
y= 2.

SIMULTANEOUS EQUATIONS. 141

EXAMPLES.
Solve by the method of addition or subtraction :

5x-\- 4:y=22. f 17ic + 10^/ = - 30.

3ic+ y = 9. [l'Sx-35y = ~m

X— 6v = — 10. .^ flla;— 5iy= 4.

^ 12. ^ -^

13. '

4a;+ 3?/ = -3. [ 8?/- 9a; = 77

6a; + 11 2/ = -28. ,, f 5a;- 9w = l.

^ 14. ' -^

5y-18a; = 8. [ 8a;-10?/ = -5

6x+ 2?/ = -3. f21a;- 8?/ = 92.

5a;- Sy = -6. ' { 9a; + 172/ = 19.

„ , 4 a; + 15?/ = 7. ,^ f 10a; - lly = -27.

8. . 16. ' ^

14a;+ 62/ = 9. [ lOy - lla; = 36

g 12a;- 52/ = 10. ^^ f 22.-^ + 152/ = 9.

30a;+ll2/ = -69. [ 18.T + 25?/ = 71

,/v , 3a;+ 7y = 2. ^ r 5a; - 24?/ = - 123.

10. ^-^ „ 18. ' ^

7a;+ 8?/ = -2. [19x - 36y = -SI.

T65, II. Elimination by Substitution.

1. Solve the equations \

7 a; — 9y= 15. (1)

2/ -5a; = -17. (2)

Transposing — 5x in (2), 8?/ = ox — 17.

Whence, y = ^^ ~ ^'^ - (3)

8

Substituting this in (1), 7x - of^^^'j =: 15.

Clearing of fractions, 56 x - 9(5 x - 17) = 120.

Expanding, 56x - 45x + 153 = 120.

1 1 X = - 33.

Whence, x = — 3.

15 _ 17

Substituting the value of x in (3), y = = - 4-

142

ALGEBRA.

KULE.

From one of the given equations find the value of one of the
unknown quantities in terms of the other, and substitute tJiis
value in place of that quantity in the other equation.

EXAMPLES.

Solve by the metliod of substitution :

3.

4.
5.
6.
7.
8.
9.

3a; + 2y = n.
4 a; + 2/ = 16.

X — 6 y = 2.

Sy- 8 a; = 29.

2x- 3y = -14.

3a;+ 7y = 48.

8x-\- 5y = 5.

3x- 2?/ = 29.

2a;+ 5?/ = 13.

7x— Ay = — 19.

3x+ 7?/ = -23.

5x-{- 4?/ = -23.

5a;+ 9y = S.

6y- 9 a; = -7.

5x+ 8y = -6.
10 a; - 12 2/ = - 5.

10.
11.
12.
13.
14.
15.
16.
17.

8a;- 3y = -6.

4 a; + 6y = 7.

r 7x+ 8y = -10.

{ll.^+ 6?/ = -19.

6x-10y = 5.
15y-Ux = -15.

9a; + 8y = -6.
12 a; + 10?/ = -7.
16x — lly = 56.
12 a;- 7?/ = 37.

7a;- 8y = -43.

5y — 6x = 35.

6 a;- 9?/ = 19.
15 a; + 7 y = — 41.

5x- 8?/ = 60.

6 a; + 7 y = — 11.

166. III. Elimination by Comparison.

1. Solve the equations

Transposing — 6?/ in (1),

Whence,

Transposing 7 y in (2),

2 a; - 5 2/ = - 16.

(1)

3x + 7y = 5.

(2)

2x-5y -16.

""- 2

(3)

3x = 5-7y.

SIMULTANEOUS EQUATIONS. 143

6-7y
3
5y- 16 _ 5 - 7 2/

Whence, x

3

Equating the values of x,

Clearing of fractions, 15 ?/ — 48 = 10 — 14 ?/.

29 ?/ = 58.
Whence, y = 2.

Substituting the value of y in (S), x= ■ ~ = — 3.

r

Rule.

From each of the given equations find the value of the same
unknown quantity in terms of the other, and place these values
equal to each other.

EXAMPLES.
Solve by the method of comparison :

2x+ y = 9. fl2:c- 6?/ = 19.

5x+ 3y = 25. [ 4y- 3a; = -11.

g x+ 2y = -2. ^^ r 6a;- 7y = -12.

4.x- 7y = 37. ■ [10 a;- 9?/ = - 12.

^ &x~ 52/ = -10. ^^ fl5.r+ 8y = -14.

5a;- 2?/ = - 17. [ 6a;+12?/ = l

lla;+ 4?/ = 3. [ 5.U+ 3?/ = 27.

8a;+ 92/ = -10. [ 8?/- 3a; = -26

7a;+ 32/ = -9. f 2.a;+ 5y = -27.

6?/- 9a; = 28. [ 11 a; -f 6r/ = -41.

12.7;-25?/ = l. r 8a;- 92/ = 6.

a;+ 4?/ = 29.

10. •!

4 .-c + 10 ?/ = — 7. [7

6a;- 5y = l. f 10a; + 18y = - 11
16.

9a; + 10y = 12. I 14y-15a; = -4.

3a;- 8?/ = -17. r 9a;- 77/ = -85.

7a;+ 6?/ = -15. " [ 4a;- 11 ?/ = - 93.

144

ALGEBRA.

MISCELLANEOUS EXAMPLES.

167. Before applying either method of elimination, each
of the given equations should be reduced to its simplest
form.

[^^ ^=0. (1)

1. Solve the equations \x + 3 ?/ + 4

[x(y-2)-y(x-5) = -lS.

(2)

From (1), 7 2/ + 28- 3x- 9 = 0, or 7y- 3x = - 19.

(3)

From (2) , xy — 2 X ■— xy + 5 y = — 13, or 5y — 2x =— 13.

(4)

Multiplying (3) by 2, Uy-6x=- 38.

(5)

Multiplying (4) by 3, 15 ?/ - 6 x = - 39.

(6)

Subtracting (5) from (6), y = - 1.

Substituting the value of y in (4), - 5 - 2 x = - 13.

-2x=- 8.

Whence, » = 4.

Solve the following:

2.

'2x Sy 7
3 4 2

X 2y_ll^
4 5 2*

■8x + 7y = 12.

x-\-2y 1 2.T + V

1

[ 4 ' 3

' x-\-5y 2y + x

13 11

dx-y = 2.

6.

x~l
5

= 0.

2a;-

3 2y

+ 13

6 + a;

-2/_

7

1-x

-y

4

2x-\-

3y = -

-1.

= 0,

3 2

3-2a; 4 + 5y^^
5 11

(x + l)(y + 9)-(a; + 5)(y-7) = 112.
2 a; + 3?/ + 9 = 0.

10

SIMULTANEOUS EQUATIONS. 145

7
x + S __ _ ^j

3

2

11.

10iK_^:=i:: = ll.

7

2 6 3 *

x-\-y—9 _ y—x~Q_t^
2 3

12.

8y

a;-y 3

3a; + ?/-3 1_^

2 ?/ - a; 11*

13.

14.

15.

16.

17.

18.

7 4

3a,_2l^=2y-4.
o

a; + y 7a;-5y3_Q

4 11

5 7

S-y 2a; + 3_y+_3,

5 4 4

1 +4y _ a; + 7 ^ ^
I 11 3 "

'|(3a^ + y)-|(2a^ + y + l)=^-

3aj-|('4x + y + |') = 5y.

3x-h2x + y + 6) = 5y.
l(3x + 2y)-h2y-x) = -8.

I

x-\-2y + 4:

\fx

9,(y.—

^'-s y + H

= 0.

!_,)_ 1(. + 2) = M.

146

ALGEBRA.

19.

x-Sy y-3x g^^

2 2

1 , 3

5 4 7

30*

20.

.8x + My = .6365.
.09 X - Ay =.1.

11

-X y

2 3^

21.

3x-^^-y=5y-l-^^-^y.
7 ^ 4

7a; + 3y + 12 o

9) — 5y — 4:

22.

2_2£ + %^^_^ 5^-2^^

17

3 x + 4 y 5 _^

3a;-42/ 13~"

23.

5a; + 6 lly — 5_-|-|
10 21 ~ *

7 a; 55y — 12 ^„
Y 25~"^'^^-

21

x-2 10 -a; y-lO^Q

5 3 4*
y-f2 _ 2^_+j _ 2_+_13 ^ ^

6 32 16 '

r2a;.y-3 4y + 5^o
25. x-^2 "^ a;-3

[(2^_32/ + l)(3a; + ll2/) + 252/2=(3» + 8?/)(2x~2/)

26.

[.322,-2.4a;-:^^5r+M^_8^_:36^+^.
.25 .5

.07a; + .l ■04y + .l^Q
.6 .3 ■

SIMULTANEOUS EQUATIONS.

147

168. Literal Simultaneous Equations.

In solving literal simultaneous equations, the method of
elimination by addition or subtraction is usually the best.

1. Solve the equations

Multiplying (1) by 6',
Multiplying (2) by b,

Subtracting,
Whence,

Multiplying (1) by a'.
Multiplying (2) by a.

Subtracting (3) from (4),
Whence,

ax + by = c.

a'x + b'y = c'.

ab'x + bbhj = b'c.
a'bx + bb'y = be'.

ab'x — a'bx = b'c — be'.

_ I'c - be' .

^ ~ ab' - a'b

aa'x + a'by = a'c.
aa'x + ab'y = ac'.

ab'y — a'by — ac' — a'c.

— 0!C^ — Qi^c
^ ~ ab' - a'b

(1)
(2)

(S)
(4)

EXAMPLES.

Solve the following :

3a; + 4y = 7a.
2a; — 5?/ = 6 6.

3.

4.

ax — by = 1.
bx 4- ccy = 1-

mx +ny =p-
7n'x -{- n'y = p'.

ax + by = m.
ex —dy = n.

(a — b)x — by = a^ — ab.
x-\-y = 2a.

x+ ay

bx + ay _
b

= —

a.

£_^ =

-1

m n

3 m 6 w

2
3

X _y _
a b

1
c

X y _
a'^b'

1
c''

10.

{a - b)x - (a + b)y = a^ + b^.
\ ay + bx — 0.

148

ALGEBRA.

11.

ax — by = 2 ah.

2bx + 2ay = 3b-- a-

12.

x~ay = b(l -Ji-ab).
bx + y — a(l + ab).

13.

'14.

15.

16.

17.

18.

19.

3 a 3b

[x — y = 2(a' — b-).

' (b ~a)x~(a — c) y = bc — a^.
(h ~ c)x — ay ~ — ac.

\b + c)x + {b - c)y = 2 ab:
{a -\- c)x — {a — c)y = 2 ac.

mx -j-ny = mn (m- + n^).
x-{-y = mn (m + w).

r.

^-by = 2b.

bx + ay --

a^b + ab'^ + 6^
ab

(a + b)x—{a — b)y = 3 ab.
(a — b) x — (a + b)y = ab.

2x—b 3x—y

a. a + 2b

2x-b _a-2y

169. Certain equations in which the unknown quantities
occur in the denominators of fractions may be readily solved
without previously clearing of fractions.

1. Solve the equations

rio

9

X

y

S ,

^^

- +

X

y

= -1.

(1)

(2)

SIMULTANEOUS EQUATIONS.

149

Multiplying (1) by 5,
Multiplying (2) by 3,

Adding,

Then,

Substituting the value of x in (1),

Then,

^-^= 40.
X y

24 45

— + — = - 3.
X y

74

= 37.

74 = 37 X, and x =2.
6-^ = 8.

y

9

= 3, and w = — 3.

y

EXAMPLES.

Solve the following :
9 10

2.

7.

8. ■!

+ — = -1.

X y

5+15=1.
x^ y

10_9_^

X y~

8_15^9^
X y 2

?>x y 9*
X 4,y 8

2x 3y 2

_2 L = ^.

3x 2y 72*

5.

2a; + - = -ll.

y

A 3 21
2

y

a b _
X y~^'

h a
- + - = c.
x y

m n
X y ^

"' y

X

hx ay

b_.a^ b*
ax by

a'b^

10.

a-i-b

X

ay

ab — b^ 1 _a^ + 3ab
X y a-\-b

11.

a + b . a

= 5b — a.

a b ^ o7>
-4-- = 2a — 3 6.
[X y

150 ALGEBRA.

XV. SIMULTANEOUS EQUATIONS.

CONTAINING MORE THAN TWO UNKNOWN
QUANTITIES.

170. If we have three independent simple equations, con-
taining three unknown quantities, we may combine any two
of them by one of the methods of elimination explained in
§§ 164 to 166, so as to obtain a single equation containing
only two unknown quantities.

We may then combine the remaining equation with either
of the other two, and obtain another equation containing
the same two unknown quantities.

By solving the two equations containing two unknown
quantities, we may obtain their values ; and substituting
them in either of the given equations, the value of the
remaining unknown quantity may be found.

We proceed in a similar manner when the number of
equations and of unknown quantities is greater than three.

The method of elimination by addition or subtraction is
usually the most convenient.

■ 6a;-4^- 7« = 17. (1)

. 9a; -7?/ -162 = 29. (2)

\fyx-by- 3« = 23. (3)

1. Solve the equations

Multiplying (1) by 3,
Multiplying (2) by 2,
Subtracting, 2 y + 11 « = - 7. (4)

Multiplying (1 ) by 5, 30 x - 20 y - 35 sr = 85. (5)

Multiplying (3) by 3, 30x - 15y - 9^ = 69. (6)

Subtracting (5) from (6), 5y + 262 = - 16. (?)

Multiplying (4) by 5,
Multiplying (7) by 2,
Subtracting, 3 « = — 3.

18 X

-12?/ -21 2 =

51.

18 X

- 14?/ -32 3 =

58.

2y-i-nz =

-7.

30 X

~20y-S6z =

85.

30 X

- 15?/- 9s =

69.

,

5y + 26z=:

-16.

10y + 55z =

-35.

10 y + 52 0 =

-32.

SIMULTANEOUS EQUATIONS.

151

z=- 1.

2y-ll=- 7.

y= 2.

6x-8 + 7= 17.

x= 3.

Whence,

Substituting in (4),
Whence,

Substituting in (1),
Whence,

In certain cases the solution may be abridged by means
of the artifice which is employed in the following example.

2. Solve the equations

Adding,
Dividing by 3,
Subtracting (2) from (5),
Subtracting (3) from (5),
Subtracting (4) from (5),
Subtracting (1) from (5),

u -\- X -\- y =: 6.
x-{-y-{-z=7.
y + z + u = 8.

.2 + M -|- CB = 9,

3m + 3x + 3?/ + 3s = 30.
u + x + y + z = 10.
m = 3.
x = 2.
y = l.
« = 4.

(1)

(3)
(4)

(6)

EXAMPLES.

Solve the following :

{3x + 2y = 13.
3y-2z = 8.
2x — 3z = 9.

(3x + 4:y + 5z = — 21.

4. \ x + y — z = — ll.
[y-8z = -20.

( 12x -4:y + z = S.

5. \x — y — 2z = — l.
l5a;-2y = 0.

7.

2x — y + z = -d.
x — 2y + z = 0.
x-y + 2z = -ll.

X — y -{- z = 9.
x-2y + 3z = 32.
x — 4:y-{-5z = 62.

3x — y-z = 7.

x-3y-z = 21.

^x-y-3z = 27.

152

ALGEBRA.

9.

10.

11.

12.

13.

14.

15.

16.

17.

2a;-32/ = 4.
4x-32 = 2.

42/ + 22 = -3.

' a;+ 2?/ — 3z = 5.
I 3x-222/ + 6z = 4.
[zx- 62/-3z = 15.

[5a;+ 7/ + 42; = -5.

3a;-52/ + 6z = -20.
[ a;_32/ — 4z = — 21.

r2x-3y-4z = -10.

3a; + 42/ + 2z = -5.
l4a;+22/ + 3z = -21.

f5x + 4?/ + 3z = 7.
I 9a;- 2/ + 6z = -39.
[8a;-72/-12z = -2.

r 2a; — 6^/ — 52 = — 11-
10a; + 9?/ -32; = 50.

[ 4a; -82/ + 2 = 15.

18.

19.

20.

21.

a; + 3T/-72; = 31.
3a; + 2/ + 52 = -49. 22.
20 .T + 2 ?/ — 5 2; = — 35.

9a; + 42/ = 10z + ll.
122/-5z = 6a;-9.
152 + 3a; = -82/ -16.

5a; + 162/ + 62 = 4.
10a; + 4y-122 = -7.
15a;-122/-3z = -10.

23.

§+ 1 = 5.
2/ a;

3.+ 3=1.

22 2/
a; 2

a; 22/ 5
2/ 32 3
2 4a; 4

x — ay — a{a^ + 1).
y — az = a\a? — 1).
2 _ ax = — a^a + 1)-

a(a; -c) +%-<')= ^•
&(2/-a)+c(2-a)=0.

[ciz-h)+a(x-h)=Q.

u-\-x \ y = 7.
x + y+z=-8.
2/ + 2 +w = 5.
2 + w + a; = — 10.

X y z

[z X y

SIMULTANEOUS EQUATIONS.

153

24.

25.

{u — 2x = -13.
x-3y = 13.

y — 4:Z —5.

z -5tt = 23.

{7x-\-4:y—3it=0.
5x-i-4:y+'iz^5u = 0.
2x+z—u=0.
2 x+4:y—3z—^l= —8.

"^ 2

z 25
3 3'

z

- - = — 2.
3

X
z

y 19

2

3 3

30.

31.

26.

r 9a; -262/ -162 = -44.

27. 12.'c-8i/ + 15z = -15.

[8a;-92/ + 13z = -24.

28.

a;

+1-

z
'3'

= 17.

X

+f=

--6

z + 2
5

•

X

z

-13

1

-2/-

—

3

2

2

32.

33.

X 2/

1

c

^-^-

1

a

. Z X

1
''b'

'^+y=

-.2b,

a c

X z
be"

:2 a.

. & a

-.2c.

b a
X y~

-.1.

a c

Z X

= 1.

[y z

= 1.

x-\-y+az = a+2.
ay-\-az + a~x=a^+a-{-l,
[ az+ax-\-ary=2a?+l.

29.

f^ + ^_? = -23.
2^3 4

3 4 2

?— ^_?=-3

4 2 3

34.

3x — yBy + A:Z_l^
5 "" 2 ~T'
2a; + 3z a;-4y^7
6 4 4*

4 .-c — z 3 1/ — 52 _ 49

154 ALGEBRA.

XVI. PROBLEMS.

INVOLVING SIMULTANEOUS EQUATIONS.

171. In solving problems where two or more letters are
used to represent unknown quantities, we' must obtain from
the conditions of the problem as many independent equations
(§ .160) as there are unknown quantities to he determined.

172. 1. Divide 81 into two parts such that three-fifths
of the greater shall exceed iive-ninths of the less by 7.

Let X = the greater part,

and y — the less.

Since the sum of the greater and less parts is 81, we have

x + y = 81. (1)

And since three-fifths of the greater exceeds five-ninths of the less

by 7,

Solving (1) and (2), x = 45, y^ 36.

2. If 3 be added to both numerator and denominator of
a fraction, its value is -f ; and if 2 be subtracted from both
numerator and denominator, its value is ^. Required the
fraction.

Let X = the numerator,

and y = the denominator.

X 4- S 2

By the conditions, ^ ' = -,

2/ + 3 3'

and ^^ = 1.

2/-2 2

Solving these equations, x = 7, y = 12.
Therefore, the fraction is ^j.

PROBLEMS. 155

PROBLEMS.

3. Divide 59 into two parts such that two-thirds of the
less shall be less by 4 than four-sevenths of the greater.

4. Find two numbers such that two-fifths of the greater
exceeds one-half of the less by 2, and four-thirds of the less
exceeds three-fourths of the greater by 1.

5. If 5 be added to the numerator of a certain fraction,
its value is | ; and if 5 be subtracted from its denominator,
its value is |. Find the fraction.

6. If 9 be added to both terms of a fraction, its value is
f; and if 7 be subtracted from both terms, its value is |.
Find the fraction.

7. A grocer can sell for $ 57 either 9 barrels of apples
and 16 barrels of flour, or 15 barrels of apples and 14 bar-
rels of flour. Find the price per barrel of the apples and
of the flour.

8. A's age is f of B's ; but in 16 years his age will be
■^ of B's. Find their ages at present.

9. If twice the greater of two numbers be divided by
the less, the quotient is 3 and the remainder 7 ; and if five
times the less be divided by the greater, the quotient is 2
and the remainder 23. Find the numbers.

10. If the numerator of a fraction be trebled, and the
denominator increased by 8, the value of the fraction is f ;
and if the denominator be halved, and the numerator de-
creased by 7, the value of the fraction is \. Find the
fraction.

11. Three years ago A's age was f of B's ; but in nine
years his age will be ^ of B's. Find their ages at present.

12. A and B can do a piece of work in 9 hours. After
working together 7 hours, B finishes the work in 5 hours.
In how many hours could each alone do the work ?

156 ALGEBRA.

13. A man invests a certain sum of money in 41 per cent
stock, and a sum $ 180 greater tlian the first in 3^ per cent
stock. The incomes from the two investments are equal.
Find the sums invested.

14. My income and assessed taxes together amount to
$ 64. If the income tax were increased one-fourth, and the
assessed tax decreased one-fifth, they would together amount
to f 63.80. Find the amount of each tax.

16. If B gives A f 12, A will have f as much money as
B ; but if A gives B $ 12, B will have ^ as much money as
A. How much money has each ?

16. A man pays with a $ 5 note two bills, one of which
is six-sevenths of the other, and receives back in change
seven times the difference of the bills. Find their amounts.

17. Find three numbers such that the first with one-third
the others, the second with one-fourth the others, and the
third with one-fifth the others may each be equal to 25.

18. A sum of money was divided equally among a cer-
tain number of persons. Had there been 3 more, each
would have received $1 less; had there been 6 fewer, each
would have received $5 more. How many persons were
there, and how much did each receive ?

Let X = the number of persons,

and y = the number of dollars received by each.

Then, xy = the number of dollars divided.

The sum of money could be divided among a; -f- 3 persons, each of
whom would receive y — I dollars ; and among x — 6 persons, each
of whom would receive y + 5 dollars.

Whence, (x + 3)(y — 1) and (x — 6)(?/ -f 5) will also represent the
number of dollars divided.

Then (x -f 3) (y - 1) = xy,

and (x — 6) (?/ -t- 5) = xy.

Solving these equations,

X = 12, y = 6.

PROBLEMS. 157

19. A man bought a certain number of eggs. If he had
bought 56 more for the same money, they would have cost
a cent apiece less ; if 24 less, a cent apiece more. How
many eggs did he buy, and at what price each ?

20. A boy spent his money for oranges. If he had got
15 more for his money, they would have cost 1^ cents each
less ; if 5 fewer, they would have cost 1^ cents each more.
How much did he spend, and how many oranges did he get ?

21. A sum of money is divided equally among a certain
number of persons. Had there been m more, each would
have received a dollars less; if n less, each would have
received b dollars more. How many persons were there,
and how much did each receive ?

22. A purse contained $6.55 in quarter-dollars and
dimes; after 6 quarters and 8 dimes had been taken out,
there remained 3 times as many quarters as dimes. How
many were there of each at first?

23. A dealer has two kinds of wine, worth 50 and 90
cents a gallon, respectively. How many gallons of each
must be taken to make a mixture of 70 gallons, worth 75
cents a gallons ?

24. A grocer bought a certain number of eggs at the rate
of 22 cents a dozen, and seven-fifths as many at the rate of
14 cents a dozen. He sold them at the rate of 20 cents a
dozen, and gained 24 cents by the transaction. How many
of each kind did he buy ?

25. A and B can do a piece of work in 10 days, A and C
'in 12 days, and B and C in 20 days. In how many days

can each of them alone do it ?

26. A resolution was adopted by a majority of 10 votes ;
but if one-foarth of those who voted for it had voted against
it, it would have been defeated by a majority of 6 v^otes.
How many voted for, and how many against it ?

158 ALGEBRA.

27. The sum of the three digits of a number is 13. II
the number, decreased by 8, be divided by the sum of its
second and third digits, the quotient is 25; and if 99 be
added to the number, the digits will be inverted. Find the
number.

Let X = the first digit,

y = the second,

and z = the third.

Then, 100 x + 10y + z = the number,

and 100 z + 10?/ + x = the number with its digits inverted.

By the conditions of the problem,

x + y + z = 13,

100a; + lOy + z -8_^^
y + 2

and lOOx + lOy + z + 99 = 100z + 10y + x.

Solving these equations, x = 2, y = S, z = 3.

Therefore, the number is 283.

28. The sum of the two digits of a number is 16 ; and if
18 be subtracted from the number, the digits will be inverted-
Find the number.

29. The sum of the three digits of a number is 23 ; and
the digit in the tens' place exceeds that in the units' place
by 3. If 198 be subtracted from the number, the digits will
be inverted. Find the number.

30. If the digits of a number of two figures be inverted,
the sum of the resulting number and twice the given num-
ber is 204 ; and if the number be divided by the sum of its
digits, the quotient is 7 and the remainder 6. Find the
number.

31. If a certain number be divided by the sum of its two
digits, the quotient is 4 and the remainder 3. If the digits
be inverted, the quotient of the resulting number increased
by 23, divided by the given number, is 2. Find the number.

PROBLEMS. 159

32. Two vessels contain mixtures of wine and water. In
one there is three times as much wine as water, and in the
other five times as much water as wine. How many gallons
must be taken from each to fill a third vessel whose capacity-
is 7 gallons, so that its contents may be half wine and half
water ?

33. If a lot of land were 6 feet longer and 5 feet wider,
it would contain 839 square feet more ; and if it were 4 feet
longer and 7 feet wider, it would contain 879 square feet
more. Find its length and width.

34. A and B are building a piece of fence 189 feet long.
After they have worked together 9 hours A leaves off, and
H finishes the work in 12|- hours. If 12 hours had occurred
before A left off, B would have finished the work in 4^ hours.
How many feet does each build in one hour ?

35. The sum of the three digits of a number is 17. The
sum of 3 times the first digit, 5 times the second, and 4
times the third is 70; and if 297 be added to the number,
the digits will be inverted. Find the number.

36. The rate of an express train is five-thirds that of
a slow train, and it travels 36 miles in 32 minutes less time
than the slow train. Find the rate of each in miles an hour.

37. Divide $ 396 among A, B, C, and D, so that A may
receive one-half the sum of the shares of B and C, B one-
third the sum of the shares of C and D, and C one-fourth
the Slim of the shares of A and D.

38. A merchant has two casks of wine, containing together
56 gallons. He pours from the first into the second as much
as the second contained at first ; he then pours from the
second into the first as much as was left in the first ; and
again from the first into the second as much as was left in
the second. There is now three-fourths as much in the first
as in the second. How many gallons did each contain at
first?

160 ALGEBRA.

39. A crew can row 10 miles in 50 minutes down stream,
and 12 miles in an liour and a half against the stream.
Find the rate in miles an hour of the current, and of the
crew in still water.

Let X = the number of miles an hour rowed by the crew in

still water,

and y = the number of miles an hour of the current.

Then, x + y = the number of miles an hour of the crew rowing
down stream,

and X — y — the number of miles an hour of the crew rowing up

stream.
Since the number of miles an hour rowed by the crew is equal to
the distance divided by the time in hours, we have

X + 2/ = 10 - ^ = 12,
b

and X - w = 12 -^ 5 = 8.

Solving these equations, x = 10, y = 2.

40. A crew can row a miles in m hours down stream, and
h miles in n hours against the stream. Find the rate in
miles an hour of the current, and of the crew in still water.

41. A vessel can go 63 miles down stream and back again
in 20 hours ; and it can go 3 miles against the current in
the same time that it goes 7 miles with it. Find its rate in
miles an hour in going, and in returning.

42. If a number of two figures, diminished by 3, be
divided by the sum of its digits, the quotient is 5. If the
digits be inverted, the quotient of the resulting number
increased by 18, divided by the sum of the digits, is 7.
Find the number.

43. The digits of a number of three figures have equal
differences in their order. If the number be divided by
one-half the sum of its digits, the quotient is 41 ; and if
594 be added to the number, the digits, will be inverted.
Find the number.

PROBLEMS. 161

44. If I were to make my field 5 feet longer and 7 feet
wider, its area would be increased by 830 square feet ; but
if I were to make its length 8 feet less, and its width 4 feet
less, its area would be diminished by 700 square feet. Find
its length and width.

^ 45. A certain sum of money at simple interest amounts
in 3 years to f 420, and in 7 years to $ 480. Required the
sum and the rate of interest.

46. A certain sum of money at simple interest amounts
in m years to a dollars, and in n years to b dollars. Re-
quired the sum and the rate of interest.

47. A and B together can do a piece of work in 8f days ;
^but if A had worked f as fast, and B f' as fast, they would

lave done it in 7|- days. In how many days could each
alone do the work ?

48. A sum of money at simple interest amounts to $ 2080
in 8 months, and to $ 2150 in 15 months. Find the sum
and the rate of interest.

49. A train running from A to B meets with an accident
which causes its speed to be reduced to one-third of what
it was before, and it is in consequence 5 hours late. If the
accident had happened 60 miles nearer B, the train would
have been only 1 hour late. Find the rate of the train
before the accident, and the distance to B from the point
of detention.

Let 3 X = the number of miles an hour of the train before the

accident.
Then, x = the number of miles an hour after the accident.

Let y = the number of miles to B from the point of detention.

V

The train would have done the last y miles of its journey in ^

hours ; but owing to the accident, it does the distance in - hours.
° X

Then, | = J^ + 6. (1)

162 ALGEBRA.

If the accident had occurred 60 miles nearer B, the distance to B
from tlie point of detention would have been y — 60 miles.

Had there been no accident, the train would have done this in

^ — hours, and the accident would have increased the time to — —
ox X

hours.

_, w - 60 w — 60

Then, L_ = l^ + 1. (2)

Subtracting (2) from (1), — =:h — 1-4, or — =4.

X o X X

Whence, x — 10.

Then the rate of the train before the accident was 30 miles an hour.

Substituting in (1), w ^ S "*" ^' ^^ Tb^^'

Whence, y — 75.

50. A train running from A to B meets with an accident
which delays it 45 minutes ; it then proceeds at five-sixths
its former rate, and arrives at B 75 minutes late. Had the
accident occurred 45 miles nearer A, the train would have
been 90 minutes late. Find the rate of the train before the
accident, and the distance to B from the point of detention.

51. The unit's digit of a number of three digits is 7. If
the digits in the hundreds' and tens' places be interchanged,
the number is decreased by 180. If the digit in the hun-
dreds' place be halved, and the other two digits interchanged,
the number is decreased by 273. Find the number.

52. A, B, C, and D play at cards, having together $ 46.
After A has won one-third of B's money, B one-fourth of
C's, and C one-fifth of D's, A, B, and C have each f 10.
How much had each at first?

53. A, B, and C have together f-24. A gives to B and
C as much as each of them has ; B gives to A and C as
much as each of them then has ; and C gives to A and B

jy as much as each of them then has. They have now equal
amounts. How much did each have at first ?

PROBLEMS. 163

54. The fore-wheel of a carriage makes 8 revolutions
more than the hind-wheel in going 180 feet; but if the cir-
cumference of the fore-wheel were |- as great, and of the
hind- wheel 4 as great, the fore-wheel would make only 5
revolutions more than the hind-wheel in going the same
distance. Find the circumference of each wheel.

55. A and B together can do a piece of work in m days,
B and C in n days, and C and A in p days. In what time
can each alone perform the work ?

56. A piece of work can be completed by A working 3
;v. days, B 7 days, and C 1 day ; by A working 5 days, B 1

)) day, and C 7 days ; or by A working 1 day, B 5 days, and
C 11 days. In how many days can each alone perform the
work ?

57. A man has a sum of money invested at a certain rate
of interest. Another man has a sum greater by $ 3000,
invested at a rate 1 per cent less, and his income is $ 45
less than that of the first. A third man has a sum less by
f 2000 than that of the first, invested at a rate 1 per cent
greater, and his income is $ 40 greater than that of the first.
Find the capital of each man, and the rate at which it is
invested.

■ 58. A and B can do a piece of work in a hours. After
working together h hours, B finishes the work in c hours.
In how many hours could each alone do the work ?

59. A crew row up stream 26 miles and down stream 35
miles in 9 hours. They then row up stream 32 miles and
down stream 28 miles in 10 hours. Find the rate in miles
an hour of the current, and of the crew in still water.

(Let X and y represent the number of miles an hour of the crew
rowing up and down stream, respectively.)

60. A sum of money, at 6 per cent interest, amounts to
$ 5900 for a certain time, and to f 7100 for a time longer
by 4 years. Find the principal and the time.

164 ALGEBRA.

61. A gives to B and C twice as mucli money as each of
them has ; B gives to A and C twice as much as each of
them then has ; and C gives to A and B twice as much as
each of them then has. Eacli has now $27. How much
did each have at first ?

62. A party at a tavern found, on paying their bill, that
had there been 4 more, each would have paid 75 cents les- •
but if there had been 4 fewer, each would have paid $ 1.50
more. How many were there, and how much did each pay ?

/ 63. An express train travels 30 miles in 27 minutes less
time than a slow train. If the rate of the express train
were f as great, and of the slow train ^ as great, the express
train would travel 30 miles in 54 minutes less time than
the slow train. Find the rate of each in miles an hour.

64. A and B run a race of 450 feet. The first heat, A
gives B a start of 135 feet, and is beaten by 4 seconds ; the
second heat, A gives B a start of 30 feet, and beats him by
3 seconds. How many feet can each run in a second ?

65. A sum of money consists of quarter-dollars, dimes,
and half-dimes. Its value is as many dimes as there are
pieces of money ; its value is also as many quarters as there
are dimes ; and the number of half-dimes is one more than
the number of dimes. Find the number of each coin.

66. A man invests $ 5100, partly in 3|- per cent stock at
$ 90 a share, and partly in 4 per cent stock at $ 120 a share,
the par value of each share being f 100. If his annual
income is $185, how many shares of each stock does he
buy?

67. A and B run a race of 336 yards. The first heat, A
gives B a start of 28 yards, and beats him by 2 seconds ;
the second heat, A gives B a start of 12 seconds, and is
beaten by 48 yards. How many yards can each run in a
second ?

mEQUALITIES. 1G5

XVII. INEQUALITIES.

173. Definitions.

The Signs of Inequality, > and < , are read " is greater
than " and " is less than,'''' respectively.

Thus, a > 6 is read " a is greater than 6 " ; a < 6 is read
" a is less than 6."

The Sign of Continuation, •••, signifies "and so on^^ or
" continued by the same law.^'

174. One number is said to be greater than another when
the remainder obtained by subtracting the second from the
first is a positive number ; and one number is said to be less
than another when the remainder obtained by subtracting
the second from the first is a negative number.

Thus, if a — 6 is a positive number, a > 6 ; and if a —b
is a negative number, a <.b.

175. An Inequality is a statement that one of two expres-
sions is greater or less than another.

The First Member of an inequality is the expression to
the left of the sign of inequality ; the Second Member is the
expression to the right of that sign.

Any term of either member of an inequality is called a
term of the inequality.

176. Two or more inequalities are said to subsist in the
same sense when the first member is the greater or the less
in both.

Thus, a > & and c>d subsist in the same sense.

177. An inequality ivill continue in the same sense after
the same quantity has been added to, or subtracted from, both
members.

166 ALGEBRA.

For consider tlie inequality a > 6.

Then by § 174, a — & is a positive number.

Hence, each of the numbers

(a -\- c) — (b -\~ c), and (a — c) — (b — c)
is positive, since each is equal to a — &.

Therefore, a + c>b + c, and a —c>b — c. (§ 174)

178. It follows from § 177 that a term may he transposed
from one member of an inequality to the other by changing its
sign.

179. // the signs of all the terms of an inequality be changed,
the sign of inequality must be reversed.

For consider the inequality a — b > c — d.
Transposing every term, we have

d-c>b-a. (§ 178)

That is, b — a<d — c.

180. An inequality will continue in the same sense after
both members have been multiplied or divided by the same
positive number.

For consider the inequality a > &.
By § 174, a — b is a positive number.
Hence, if m is a positive number, each of the numbers

a~b

m{a — h) and

m

or, ma — mb and — - — , is positive

m m '^

Therefore, ma'^mb, and — > —

mm

181. It follows from §§ 179 and 180 that if both members
of an inequality be multiplied or divided by the same negative
number, the sign of inequality must be reversed.

f

INEQUALITIES. 167

182. If any number of inequalities, subsisting in the same
sense, be added member to member, the residtirig inequality
will also subsist in the same sense.

For consider the inequalities a>b, a' > b', a" > b", •••.
Then each of the numbers a — b, a' — b', a" — b", •••, is
positive.

Therefore, their sum

a-b + a'-b' + a" -b" ^-—,

or, a + a' + a"H (b + b' -\- b" -\ ),

is a positive number.

Whence, a + a' + a" -\ > & + 6' + 6" -1 .

183. It is to be observed that, if two inequalities, subsist-
ing in the same sense, be subtracted member from member,
the resulting inequality does not necessarily subsist in the
same sense.

Thus, if a > 6 and a' > 6', the numbers a—b and a'—b'
are positive.

But (a — 6) — (a' — b'), or its equal (a— a')— (6 — b'), may
be positive, negative, or zero ; and hence a— a' may be greater
than, less than, or equal to b — b'.

EXAMPLES.

184. 1. Find the limit of x in the inequality

Multiplying both members by 3 (§ 180), we have

21x-23<2x + 15.
Transposing (§ 178), and uniting terms,

19x<38.
Dividing both members by 19 (§ 180),

X < 2, Ans.

168 ALGEBRA.

2. Find the limits of x and y in the following:

3

x-{-2y>37

(1)

.2

x + 3y = 33

(2)

Multiplying (1) by 3,

9x + 6y>llh

Multiplying (2) by 2,

4x + 6y = 66.

Subtracting (§177),

5x> 45.

Whence,

x>9.

Multiplying (1) by 2,

6x + iy> 74.

Multiplying (2) by 3,

6x + 9y-- 99.

Subtracting,

-5y>-25.

Dividing both members

by

- 5 (§ 181),

y<6.

Find the limits of x in the following :

3. {6x-iy-28<{4.x-3){9x + 2).

4. (3a; + 2)(4a;-5)>(2i^-3)(6x + l) + 5.

5. (5aj + 1)2 + 15 > (3a;- 2)2+ (4a; + 3)2.

6. (x -2){x- 3) (a; + 4) < (a; + 1) (x + 2){x- 4).

7. 6 mx — 5 an > 15 am — 2 nx, if 3 m + ?i is negative,

8. — ^t Il_ < 2, if a and b are positive, and a>b.

a b

Find the limits of x and y in the following :

g |4a; + 9y<40. jQ |7a; + 22/>25.

l6a;-y = 2. ' \3x + 5y = 19.

11. Find the limits of x when

5a; + 7<9a;-13, and 11 a; - 20 < 6a; + 25.

12. A certain positive integer, plus 21, is greater than 8
times the number, minus 35 ; and twice the number, plus
11, is less than 7 times the number, minus 19. Find the
number.

INEQUALITIES. 169

13. A teacher has a number of his pupils such that 8 times
their number, minus 31, is less than 3 times their number,
plus 69 ; and 13 times their number, minus 45, is greater than
7 times their number, plus 57. How many pupils has he ?

14. A shepherd has a number of sheep such that 4 times
the number, minus 7, is greater than 6 times the number,
minus 89 ; and 5 times the number, plus 3, is greater than
twice the number, plus 114. How many sheep has he ?

15. Prove that if a and b are unequal positive numbers,

0 a

Since the square of any number is positive,

(a-6)2>0.
That is, a2 - 2 a6 + 62 > 0.

Transposing —2ab, a^ + b'^>2 ah.

Dividing each term of the inequality by ab (§ 180), we have

h a

16. Prove that for any value of x, except 1, a;^ + 1 > 2 x.

17. Prove that for any value of x, except |, 9 a^ + 4 > 12 x.

■ In each of the following examples, the letters are under-
stood as representing positive numbers.

18. Prove that -^ + — > 2, if 6 is not equal to i a.

2b a

19. Prove that (a + 2 6) (a - 2 &)> & (6 a - 13 b), if b is
not equal to ^a.

20. Prove that a(9a - 4&) >46(2a - &), if b is not
equal to fa.

21. Prove that (a" -b'){c'- dF) < {ac - bdf, if 6c does
not equal ad.

170 ALGEBRA.

XVIII. INVOLUTION.

185. Involution is the process of raising a given expres-
sion to any required power whose exponent is a positive
integer.

This may be effected, as is evident from § 6, by taking
the expression as a factor as many times as there are units
in the exponent of the required power.

INVOLUTION OF MONOMIALS.

186. 1. Find the value of (5 a^b-f.

By § 6, (5 a^b^y = 5 a^b'^ x 5 a^'^ x 5 a^b^ = 125 a^b^, A7is.

2. Find the value of (- ay.

(-«)*= (_a)x(-a)x(- a)x(- a)=a* (§ 49), Ans.

3. Find the value of (- 3 ony.

( _ 3 to4)3 _ ( _ 3 „i4) X ( - 3 TO*) X ( - 3 m*) = - 27 m^^ (§ 49) , Ans.

From the above examples, we derive the following rule :

Raise the absolute value of the numerical coefficient to the
required power, and multiply the exponent of each letter by the
exponent of the required power.

Give to every power of a positive term, and to evenly Even
power of a negative term, the positive sign, and to every Odd
power of a negative term the negative sign.

EXAMPLES.
Find the values of the following :

4. (a'b^cy. 8. (2mhiy. 12. (pq'^r*y\

5. {x'^y'zY. 9. {-a^b'^cy. 13. (-GxYz'y.

6. (-m^npy. 10. (x'^yzy. 14. (2a''x'y.

7. {- 12 a'^'b^y. 11. i-Sx'y^^y. 15. (-5 7nnyy.

INVOLUTION. 171

A fraction may be raised to any required power by rais-
ing both numerator and denominator to the required power.
and dividing the first result by the second.

16. Find tlie value of [ - ^^^ ) •

V 3 2/2 ; (3 2/2)4 81J/8'
Find the values of the following :

8aV V^''^V V4a«6«

SQUARE OF A POLYNOMIAL.
187. We find by multiplication :

a +b -\- G
a -i-b -i- G

a^ + ab -{- ac

+ ab +b^-{- be

+ ac + bc-{- c^

a2 + 2a6 4- 2ac + 62 4- 2 6c + c2
The result, for convenience of enunciation, may be written :

(a 4- 6 + c)2 = a^ 4- 6- + c2 + 2 a6 + 2 ac + 2 6c.
In like manner we find :
(a + 6 + c + d)2 = a^ + 6^ + c2 + ^2

+ 2a6 + 2ac + 2ad + 26c + 26d+2cc?;
and so on.

We then have the following rule :

The square of a polynomial is equal to the sum of the squares
of its terms, plus twice the product of each term by each of the
following terms.

172 ALGEBRA.

EXAMPLES.

1, Square 2x^ — 3x — 5.

The squares of the terms are 4x*, 9x^, and 25.
Twice the first term into each of the following terms gives the
results — 12 x^ and — 20 x-.

Twice the second term into the following term gives the result 30 x.

Then, (2«2 _ 3x - 5)2 = 4x4 + 9x' + 25 - 12x3 - 20x2 + 30x
= 4x*-12x3-llx2 + 30x + 25, ^ns.

Square each of the following :

2. a-b-c. 11. a2-4a6 + 3 6-.

3. x-y + z. 12. 2a^ + 3xy + y-.
4:. x^ + 2x + l. 13. ar' + 6a;--7.

5. m + 2n~3p. 14. 4a*-5aV-3x«.

6. 2 a^ — a + 4. 15. a — b — c — d.

7. 5a;2-3a;-l. 16. a + b-c-^d.

8. 3a^ + 4a; + 2. 17. .^^ - a^2 _ a; + 2.

9. 6^3 + n- 5. 18. a3 + 2a2-3a-4.
10. 2a- 56- c. 19. 2a;3_5a^^4^_3

CUBE OF A BINOMIAL.

188. We find by multiplication :

(a + by = a' + 2ab + b^
a +b

a* + 2a2&+ ab^

a-b + 2 a&^ + b^
(a + 6)3 = a3 ^ 3 a^^ + 3 ad^ + 6'
(a -by = d'-2ab + 6^

a — 6

-2a26+ a6'

- a-6 + 2 ab- - ft^

(a - 6)3 = a3 _ 3 a==6 + 3 a62 - 6^

INVOLUTION. 173

That is, the cube of the sum of two quantities is equal to
the cube (f the first, plus three times the square of the first
times the second, jylus three times the first times the square oj
the second, plus the cube of the second.

The cube of the difference of two quantities is equal to
the cube of the first, minus three times the square of the first
times the second, plus three times the first times the square
f the second, minus the cube of the second.

EXAMPLES.

1. Find the cube of a + 2 &.

We have, (a + 2 6)3 = a^ + 3 a^-(2 6) + 3 a(2 &)2 + (2 by
= a3 + 6 a^b + 12 ab'^ + 8 b^ Ans.

2. Find the cube of 2x^-3 y^

(2 x3 - 3 ?/2)3 = (2 x3)3 -3(2 r'')2(3 2/2) +3(2 x^) (3 y^y - (3 2/2)3
= 8 x^ - 36 x«2/2 + 54 x^y* - 27 y^, Ans.
Cube each of the following :

3. a.t + by. 7. «- + 5. 11. 3x- — hx.

4. a; + 2. 8. 6a -6. 12. ^x^ + byzK

5. 3a-l. 9. 5x + 2y. 13. 2a;-7ar'.

6. m-4n. 10. 4m -Sn^'. 14. 5a« + 6 6^.
The' cube of a trinomial may be found by the above

method, if two of its terms be enclosed in a parenthesis
and regarded as a single term.

15. Find the cube of a^ — 2 a; — 1,

(x2-2x-l)3 = [(x2-2x)- 1]3

= (x2 - 2x)3 - 3(x2 - 2x)2 + 3(x2 - 2x)- 1
= x6-6x5+12x*-8x3-3(x*-4x3+4x2) + 3(x2-2x)-l
= x6-6x5+12x*-8x3-3x* + 12x3-12x2+3x2-6x-l
= x6 - 6 x5 + 9 X* + 4 x3 - 9 x2 - 6 X - 1, Ans.

Cube each of the following :

16. a + b-c. 18. x-y + 2z. 20. 2x'' + x-3.

17. x'^x + l. 19. a'-Sa-l. 21. 3-4x- + a^.

V

174 ALGEBRA.

XIX. EVOLUTION.

189. If an expression when raised to the nth power, n
being a positive integer, is equal to another expression, the
first expression is said to be the ?ith Root of the second.

Thus, if a" = b, a is the 7ith root of b.

190. Evolution is the process of finding any required
root of an expression.

191. The Radical Sign, -^^, when written before an ex-
pression, indicates some root of the expression.

Thus, Va indicates the second, or square root of a ;
■\/a indicates the third, or cube root of a ;
■y/a indicates the fourth root of a ; and so on.

The index of a root is the number written over the radical
sign to indicate what root of the expression is taken.
If no index is expressed, the index 2 is understood.

EVOLUTION OF MONOMIALS.

192. 1. Required the cube root of a^6V.

We have, (ab^c^)^ = a^b^c^.

Whence, Va^b^^ = abH"^. (§ 189)

2. Required the fifth root of — 32 aP.
We have, ( - 2 a)^ = - 32 a^.

. Whence, y/-Z2a^ = - 2 a.

3. Required the fourth root of a*.

We have either (+a)*or(— a)* equal to a*.
Whence, Va* — ±a.

The sign ±, called the double sign, is prefixed to an ex-
pression when we wish to indicate that it is either -f or — .

EVOLUTION. 175

193. From § 192 we derive the following rule :
Extract the required root of the absohite value of the numeri-
cal coefficient, and divide the exponent of each letter by the
index of the required root.

Give to every even root of a positive term the sign ± , and
to every odd root of any term the sign of the term itself

EXAMPLES.
1. Find the square root of 9a*6^c^°.

By the rule, V9¥¥c^ = ±3 aWd', Ans.

2. Find the cube root of — 64 x^y^'^.

V— G4a;'^j/3'' = — 4 a;^?/", Ans.
Find the values of the following :

3. V49a«62. 9. ^/^yV«.

4. -v/125xy. 10. \/243a2^.

5. -v/-m%y^ 11. ^/-512m%V8.

6. -v/lGa^m^. 12. Vl44a*'a^'+«.

7. V64^V^*- 13. V^^lWs^^^^^Sf^.

8. ^-a^ftisc^. 14. \/256a8"6i^

To find any root of a fraction, extract the required root of
both numerator and denominator, and divide the first result
by the second.

15. Find the value of ^l-'^l^.

We have, ^fl^T^ = - <^M^ = -^-^, Ans.

Find the values of the following :

64 m^

17 ^'lE?^. 19 ,^r^y 21 ^'/^

176

ALGEBRA.

The root of a large number may sometimes be found by
resolving it into its prime factors.

22. Find the square root of 254016.

We have, V254016 =^ V'26 x 3* x 7^ = 2^ x 32 x 7 = 504, Ans.
23. Find the value of -^72 x 75 x 135.

We have, V72 x 75 x 135 = V(23 x 32) x (3 x 52) x (33 x 5)

= \/23 x 36 X 53 = 2 X 32 X 5 = 90, Ans
Find the values of the follovdng :

24. V3l36. 26. V63o04.

28. V42 X 56 X 147.

25. Vi8225. 27. V48 x 54 x 72. 29. Vi3824.

30. Vl5a6 X 216c X 35 ca.
32. v9n25. 33. a/20736.
35. \/63 X 162 X 196.

31. V213444.

34. a/7776.

36. a/56 X 98 X 112.

37. Viar + 5a + 6){a' + 2a - S^a' + a - 2).

^SQUARE ROOT OF A POLYNOMIAL.

194. Since (a + b)- = a^ + 2 a6 + b-, we know that the
square root of a- +2ab + b- is a + b.

It is required to find a process by which, when the ex-
pression a^ + 2 ah + b^ is given, its square root may be
determined.

a- + 2 a& + 52

«2

2a + b

a + 6 The first term of the root, a, is found
by taking the square root of the first
term of the given expression.

Subtracting the square of a from
the given expression, the remainder is
2ab + ft2, or (2a + b)b.
If we divide the first term of this remainder by 2 a, that is, by twice
the first term of the root, we obtain the second term of the root, b.

2ab + b^
2ab + b^

EVOLUTION. 177

Adding this to 2 a, we obtain the complete divisor, 2a + b.
Multiplying this by b, and subtracting the product, 2 ab + b"^, from
the remainder, there are no terms remaining.

From the above process, we derive the following rule :

Arrange the expression according to the powers of some
letter.

Extract the square root of the first term, write the resxdt as
the first term of the root, and subtract its square from the given
expression, arranging the remainder in the same order of pow-
ers as the given expression.

Divide the first term of the remainder by twice the first term
of the root, and add the quotient to the part of the root already
found, and also to the trial-divisor.

Multiply the comjilete divisor by the term of the root last
obtained, and subtract the product from the remainder.

If other terms remain, proceed as before, doubling the part
of the root already found for the next trial-divisor.

EXAMPLES.

195. 1. Find the square root of 9 a;* - 30 a^x"^ + 25 a«.

9 X* - 30 a3x2 + 25 aS 3 x^ - 5 a^, Ans.
9x*

6x^-5a3

- 30 a3x2 + 25 ojs

- 30 a^x^ + 25 a8

The first term of the root is the square root of 9x* or Sx^.

Subtracting the square of 3x-, or9x*, from the given expression,
the first term of the remainder is — 30 a^x^.

Dividing this by twice the first term of the root, or 6 x^, we obtain
the second term of the root, — 5 a^.

Adding this to 6 x^, we have the complete divisor, 6 x^ _ 5 ^3,

Multiplying this complete divisor by — 5 a^, and subtracting the
product from the remainder, there is no remainder.

Hence, 8 x^ — 5 a^ is the required square root.

2. Find the square root of

12 a;^ - 22x3 + 1 - 20 a;< + 9a;« + 8a; + 12 a;2^

L^-

■^.A-j^M-

'y.

178

ALGEBRA.

Arranging according to the descending powers of x, we have :

3x3+2x-— 4x— 1, -4ns.

9x6 + 12a
9x6

5-20 :i

*-22x3+12x2 + 8x + l

6x3+2x2

12x6
12x5+ 4x*

6x3+4x2-4x

-24x*
-24x*-16x3 + 16x2

6x3+4

x2-

3x-l

- 6x8- 4a;2

- 6x3- 4x2+8x+l

It will be observed that eacli trial divisor is equal to the
preceding complete divisor, ivith its last term doubled.

To avoid needless repetition, the last five terms of the
first remainder, the last four terms of the second remainder,
and the last two terms of the third remainder are omitted.

Note. Since every square root has the double sign (§ 192), the
result may be written in a different form by changing the sign of eacli
term. ;

Thus, in Ex. 2, the answer may be written 1 + 4 x — 2 x^ — 3 x^. i

Find the square roots of the following :

3. 4:X* + 4:X^ + 5af + 2x + l.

4. l-6a + lla^-6a^ + a*.

5. 9a;^-24a^+4ar'+16a; + 4.

6. 20a^-70a; + 4ic* + 49-3a;2.

7. a^ + 6^ -I- c^ - 2 «& - 2 ac + 2 be.

8. 9a*4-l-4a^ + 4a«-6a2+-12a^

9. a;« - 4 x^a^ + 10 a^a^ + 4 x^a* - 20 xa' +- 25 a«

10. 9a^ + 25/+162' + 30a;2/-24iC2-40?/z.

11. 49 m* — 14 m^n — 55 mhi^ + 8 mn^ + 16 n\

12. 49 a^ _ 30 a^ 4-16 + 9 a* -40 a.

13. 25 X* - 20 x^y - 26 .vY -{- 12 xf + 9 y*.

14. 16 m* + 8 m^x^ - 23 m^x* - 6 7nx^ + 9 a^.

EVOLUTION. 179

15. 20 ab' + 9 a^ - 26 a'b' + 25 b* - 12 a'b.

16 4

16. m' + 8 m + 12 - - + — ,-

m m~

17. i_2a; + 3.T2-4.x^-f3a;*-2ay' + a;<'.

18. 12 .ij^ + 12a; - 8.^'^ + 9 + 28a;2 + a;« + 10 ar'.

' 19- ^-^'^— /+2f + 4|-

^ a;' a^ ■ Sla;'' 2.t 4

■ 9 3 60 5 25'

21. 4 a«+ 12 a'b + 25 aV-+ 4 a%'~ 14 a^i*- 40 a6''+ 25 6«

4^3"^ 36 "^6 "^16*
23. 28 xY + 9x^- 15 are/* - 12 x'y -2,xf-2 x'^f + \^f.

„. 16 , 8a; 13 a.-^ 4.^3 , 4x*
9 3 a Scv cr cr

Find to four terms the approximate square roots of:

25. 1 + 4a;. 27. 1 - a;. 29. ar' + 6.

26. l+2a. 28. l-3a. 30. 4a2-2&.

SQUARE ROOT OF AN ARITHMETICAL NUMBER.

196. The square root of 100 is 10 ; of 10000 is 100 ; etc.

Hence, the square root of a number between 1 and 100 is
between 1 and 10 ; the square root of a number between 100
and 10000 is between 10 and 100 ; etc.

That is, the integral part of the square root of a number
of one or two figures, contains one figure ; of a number of
three or four figures, contains two figures; and so on.

Hence, if a point be placed over every second figure of any
integral number, beginning with the units' place, the number
of points shows the number of figures in the integral part of
its square root.

or = 3600

120 + 8
= 2a + b

180 ALGEBRA.

197. Let it be required to find the square root of 4624.

a- + 2a& + 6- = 4624 60 + 8 Pointing the number ac

_ I 7 cording to the rule of § 196,
ri 1 ,o we find that there are two
^^ figures in the integral part of

its square root.

Let a denote the greatest
multiple of 10 whose square is less than 4624 : this we find by inspec-
tion to be 60.

Let h denote the digit in the units' place of the root; then, the
given number is denoted by (a -|- hf, or a^ j^2ab + b'\

Subtracting a^, or 3600, from 4624, the remainder is 1024.

That is, 2ab + b^- = 1024. (1)

Since &' is generally small in comparison with 2 ab, we may obtain
an approximate value of b by neglecting the b^ term in (1).

Then, 2ab = 1024, and b = ^^ = 1^ = 8 +.

2 a 120

This suggests that the digit in the units' place is 8.
If this be correct, 2ab + b~, or (2 a-)- b)b, must equal 1024.
Adding 8 to 120, multiplying the sum by 8, and subtracting the
product from 1024, there is no remainder.

Hence, 60 + 8, or 68, is the required square root.

Omitting the ciphers, for the sake of brevity, and con-
densing the operation, it will stand as follows :

4624
36

68

128

1024
1024

From the above example, we derive the following rule :

Separate the number into periods by pointing every second
figure, beginning ivith the units' place.

Find the greatest square in the left-hand period, and lorite
its square root as the first figure of the root; subtract the
square of the first rootfigure from the left-hand period, and to
the result annex the next period.

EVOLUTION.

181

Divide this remainder, omitting the last figure, by twice the
part of the root alreadij found, and annex the quotient to the
root, and also to the trial-divisor.

Multiply the complete divisor by the rootfigure last obtained,
and subtract the product from the remainder.

If other periods remain, jjroceed as before, doubling the part
of the root already found for the next trial-divisor.

Note 1. It sometimes happens that, on multiplying a complete
divisor by the figure of the root last obtained, the product is greater
than the remainder.

In such a case, the figure of the root last obtained is too great, and
one less must be substituted for it.

Note 2. If any root-figure is 0, annex 0 to the trial-divisor, and
annex to the remainder the next period.

198. Required the square root of 4944.9024.

\ 10000 VlOOOO

7032

49449024
49

1403

4490
4209

14062

28124
28124

Since 14 is not contained in 4, we write 0 as the second root-figure,
annex 0 to the trial-divisor 14, and annex to the remainder the next
period, 90. (See Note 2, § 197.)

Then,

V4944.9024 = ^^ = 70.32.
100

The work may be arranged as follows :

4944.9024
49

70.32

1403

44 90
42 09

14062

2 8124
2 8124

182

ALGEBRA.

. It follows from the above that, if a point he placed over
every second figure of any number, beginning ivith the units'
place, and extending in either direction, the rule of § 197 may
be applied to the result and the decimal point inserted in Up,
proper position in the root.

EXAMPLES.
199. Find the square roots of the following :

1. 4225.

2. 21904.

3. 508369.

4. 65.1249.

5. .156816.

6. .064516.

7. 3956.41.

8. 96.4324.

9. .00321489.
10. 12823561.

11. 75570.01.

12. .16216729.

13. 2666.6896.

14. .0062504836.

15. 86.825124.

If there is a final remainder, the number has no exact
square root ; but we may continue the operation by annex-
ing periods of ciphers, and thus obtain an approximate
root, correct to any desired number of decimal places.

16. Find the square root of 12 to four decimal places.
12.06000606 I 3.4641 +, Ans.

64

3 00
2 56

686

4400
4116

6924

28400

27696

69281 I 70400

Find the first five figures pf the square root of:
17. 7. 20. 13. '''^-23. .2. 26. .009.

18. 8.

21. 48.

24.

.056.

"•27. .00074.

19. 10.

22. 64.7.

-25.

.39.

28. 8.5645.

EVOLUTION. 183

The square root of a fraction may be obtained by taking
the square root of the numerator, and then of the denomi-
nator, and dividing the first result by the second.

If the denominator is not a perfect square, it is better to
reduce the fraction to an equivalent fraction whose denomi'
nator is a perfect square.

29. Find tte value of a/- to five decimal places.

>8 \l6 Vl6 4

Find the first four figures of the square root of:

1^. 3..

32 12

^. 39. 1^.

14 27

30. §. 32. — .

'34.

3
5

31. f. 33. I

35.

7

— •

8

36. — . 38. ^^

CUBE ROOT OP A POLYNOMIAL.

200. Since (a + 6)« = a^ + 3 a^b + 3ab^ + b% we know that
the cube root of a^ -t- 3 a-b + 3 ab^ + &^ is a + &.

It is required to find a process by which, when the expres-
sion a^ + 3a^b + 3 ab^ + b^ is given, its cube root may be
determined.

a^ + 3a^b + 3ab^ + b^\ a+b

Sa' + Sab + b^

3 a% + 3 a62 + &»
3a^b + 3ab^ + l^

The first term of the root, a, is found by taking the cube root of
the first term of the given expression.

Subtracting the cube of a from the given expression, the remainder
is 3 a25 + 3 3^,2 + 53^ or (3 a'^ + 3 ab + h'^)b.

If we divide the first term of tliis remainder by 3 a^, that is, by
three times the square of the first term of the root, we obtain the
second term of the root, h.

184 ALGEBKA.

Adding to the trial-divisor 3 a6, that is, three times the product of
the first term of the root by the second, and 6^^ that is, the square
of the second term of the root, we obtain the complete divisor,
3 a2 + 3 a6 + 62,

Multiplying this by ?>, and subtracting the product, Sa-b + Sab'^+b^t
from the remainder, there are no terms remaining.

From the above process, we derive the following rule :

Arrange the expression according to the poivers of some
letter.

Extract the cube root of the first term, write the result as the
first term of the root, and subtract its cube from the given
expression ; arranging the remainder in the same order of
powers as the given expression.

Divide the first term of the remainder by three times the
square of the first term of the root, and write the result as the
next term of the root.

Add to the trial-divisor three time's the product of the term
of the root last obtained by the part of the root previously
found, and the square of the term of the root last obtained.

Multiply the comjylete divisor by the term of the root last
obtained, and subtract the 2'>roduct from the remainder.

If other terms remain, jwoceed as before, taking three times
the sqitare of the part of the root already found for the next
trial-divisor.

EXAMPLES.
201. 1. Find the cube root of

8a^ - 36a;V + 54a^/ - 272/^.

8«6 _ 36a;«2/ + b^x^y"^ -27 y^ 2x^-3y, Ans.

8«6

12a;*-18a;2y + 9 2/2

- 36 x*y + 54 x^y^ - 27 y^
-36x^j/ + 54x2y2-27?/3

The first term of the root is the cube root of 8 x^, or 2 a;2.
Subtracting the cube of 2x^, or Sx^, from the given expression,
the first term of the remainder is — 36 x*y.

EVOLUTION.

185

Dividing this by three tniies the square of the first term of the root,
or 12 a;*, we obtain the second term of the root, — Sy.

Adding to the trial-divisor thvee times the product of the term of
the root last obtained by the part of the root previously found, or
— I8x-y, and the square of the term of tlie root last obtained, or 9y^,
we have the complete divisor, 12 x* — I8x-y + dy"^.

Multiplying this complete divisor by —3;/, and subtracting the
product from the remainder, there is no remainder.

Hence, 2 x'^ — 3 ?/ is the required cube root.

2. Find the cube root of

Arranging according to the descending powers of x, we have

x6_6x5+ 3xH28x3-9x2-54x-27 x2-2x-3,

x^

3x*-6x3+4x2

— 6x5

-6x5 + 12x4- 8x3

3x4-12x3+ 12x2

- 9x2+18x + 9

3x*-12x3+ 3x2+18x+9

- 9x4 + 36x3

- 9x4+36x3-9x2-54x-27

The second complete divisor is formed as follows :

The trial-divisor is three times the square of the part of the root
already found; that is, 3(x- — 2x)2, or 3x4 — 12x3 + 12x2.

Three times the product of the term of the root last obtained by the
part of the root previously found is 3(— S)(x^ — 2x), or — 9x2 + 18 x.

The square of the term of the root last obtained is (— 3)2, or 9.

Adding these, the complete divisor is 3 x4 — 12 x^ + 3 x2 + 18 x + 9.

The last five terms of the first remainder and the last
three terms of the second remainder are omitted.

Find the cube roots of the following:

3. 8cc^ + 12x^ + 6x+l.

4. 1 - 12 a'^ -f 48 a*' - 64 al

5. 27 m« -f- 135 m'n + 225 7nhi^ + 125 n\

6. 294 ab^ - 84 a'b - 343 b^ + 8 al

7. x^-6:t^ + 9x* + 4:a^-9x'-6x-l.

186 ALGEBRA.

8. 8a« + 36a5 + 66a* + 63a3 + 33a2 + 9a + l.

9. 30/ + 27 / + 12 ?/ - 45^* - 8 - 351/3 + ^i y\

10 ^ — ^'^ 4- ^ _ i^
■ 8"~T ~6 27*

11. 9a3-36a + a^+21a*-9a5-8-42a2.

12. 174 aj* + 8 + 1T4 ar' - 60 a;^ - 245 o^ + 8 a;*' - 60 a;.

13. 27 a^- 54 a^6 + 63 a*h''- 44 a^63+ 21 a?h^- 6 a&^+ 6«.

14. 6 a.-^?/ + 96 a;/ + 56 a,-^/ ^_ ^^e 4. 24 a;*^^ ^ 54 ^e ^ gg ^^4^

IK a^ a^ , 7aj ^ , 28 48 , 64
27 3 3 a; a^ ^ a;^

CUBE ROOT OF AH ARITHMETICAL NUMBER.

202. The cube root of 1000 is 10 ; of 1000000 is 100 ; etc.
Hence, the cube root of a number between 1 and 1000 is

between 1 and 10 ; the cube root of a number between 1000
and 1000000 is between 10 and 100 ; etc.

That is, the integral part of the cube root of a number of
one, two, or three figures, contains one figure ; of a number
of four, five, or six figures, contains two figures ; and so on.

Hence, if a point he placed over every third figure of any
integral number, beginning ivith the units' place, the number
of points shows the number of figures in the integral part of its
cube root.

203. Let it be required to find the cube root of 157464.

a^ + 3 a^d + 3 ab^ + h^= 157464
a^ = 125000

50 -f 4 = a + &

3 a- = 7500

^ab= 600

b''= 16

3 a^ + 3 a6 + &2 = 8116

32464 = 3 a^fi + 3 ab"" + 6^

32464

Pointing the number according to the rule of § 202, we find that
there are two figures in the integral part of its cube root.

EVOLUTIOJS. 187

Let a denote the greatest multiple of 10 whose cube is less than
157464; this we find, by inspection, to be 50.

Let 6 denote the digit in the units' place of the root ; then, the
given number is denoted by (a + h)^, or a* + 3 a-b + 3 ab- + b^.

Subtracting a^, or 125000, from 157464, the remainder is 32464.

That is, Za-b + ^ab'^+b^ = Z2\U. (1)

Since 3 ab"^ and b^ are generally small in comparison with 3 a%, we
may obtain an approximate value of b by neglecting the 3 ah- and b^
terms in (1).

Then, 3 a% = 32464, and ?> = §2464 ^ 32464 ^

3a2 7500

This suggests that the digit in the units' place is 4.

If this be correct, 3 a-b + 3 db"^ + 6^ or (3 d^ + 3 a6 + b'^)b, must
equal 32464.

Adding to 7500 3 ab, or 600, and h'^, or 16, the sum is 8116 ; multi-
plying this by 4, and subtracting the product from 32464, there is no
remainder.

Hence, 50 + 4, or 54, is the required cube root.

Omitting the ciphers for the sake of brevity, and con-
densing the operation, it will stand as follows :

157464
125

54

7500

600

16

8116

32464

32464

From the above example, we derive the following rule :

Separate the number into periods by pointing every third
figure, beginning with the units' place.

Find the greatest cube in the left-hand period, and write its
cube root as the first figure of the root; subtract the cube of the
first root-figure from the left-hand period, and to the result
annex the next j^eriod.

Divide this remainder by three times the square of the part
of the root already found, ivith two ciphers annexed, and write
the quotient as the next figure of the root.

188

ALGEBRA.

Add to the trial-divisor three times the product of the last
root-Jigure by the part of the root previously found, with one
cipher annexed, and the square of the last rootfigure.

Multiply the comp>lete divisor by the figure of the root last
obtained, and subtract the product from the remainder.

If other periods remain, proceed as before, taking three
times the square of the part of the root already found, ivith two
ciphers annexed, as the next trial-divisor.

Note 1. Note 1, p. 181, applies with equal force to the above rule.

Note 2. If any root-figure is 0, annex two ciphers to the trial-
divisor, and annex to the remainder the next period.

204. If, in the example of § 203, there had been more
periods in the given number, the next trial-divisor would
have been three times the square of a+b, or 3a^+6a6 + 36^.

We observe that this may be obtained from the preceding
complete divisor, Sar + oab + 6^, by adding to it its second
term, 3 ab, and twice its third term, 2 b^.

Hence, if the first number and twice the second number
required to comj)lete any trial-divisor, be added to the comr
plete divisor, the result, toith two ciphers annexed, will be the
next trial-divisor.

205. Required the cube root of 8144.865728.
We have, \/8 144.865728 = Y

/8144865728 _ V814486r)728
1000000 ~ ^1000000 '
8144865728 2012

120000
600

1

120601

600

2

12120300
12060

4

12132364

144865

120601

24264728

24264728

EVOLUTION.

189

Since 1200 is not contained in 144, the second root-figure is 0 ; we
then annex two ciphers to the trial-divisor 1200, and annex to the
remainder tlie next period, 8(i5.

The second trial-divisor is formed by the rule of § 204. Adding to
the complete divisor 120601 the first number, 000, and twice the second
number, 2, required to complete the trial-divisor 120000, we have 121203 ;
annexing two ciphers to this, the result is 12120o00.

Then,

V'8144.865728 = ?^ = 20. 12.
100

The work may be arranged as follows :

8144.865728

8

120000

144 865

600

1

120601

120 601

600

24 264728

2

12120300

12060

4

12132364

24 264728

20.12

It follows from the above that, if a point he placed over
every third figure of any number, beginning with the units'
place, and extending in either direction, the rule of § 203
may be applied to the result, and the decimal point inserted
in its proper position in the root.

EXAMPLES.
206. Find the cube roots of the following

/

1. 19683.

2. 148877.

3. 59.319.

4. .614125.

5. 2515456.

6. 857.375.

7. .224755712.

8. 46.268279.

9. 523606616.
10. 187149.248.

11. .000111284641.

12. 785^889.024.

13. 444.194947.
14. 338608873.

15. .001151022592.

190 ALGEBRA.

. Find the first four figures of the cube root of :

16. 3. 18. 9.1. 20. I 22. 1

17. 7. 19. .02. 21. ||. 23. |.

207. If the index of the required root is the product of
two or more numbers, we may obtain the result by successive
extractions of the simpler roots.

For by § 189, fVa)""' = a.

Taking the nth root of both members,

f-v/ar^V^. (1)

Taking the mth root of both members of (1),

Hence, the mrith root of an expression is equal to the mth
root of the nth root of the expression.

Thus, the fourth root is the square root of the square
root ; the sixth root is the cube root of the square root, etc.

EXAMPLES.
Find the fourth roots of the following :

1. 81 a* + 216 a^h^ + 216 a'h' + 96 a&« + 16 h\

2. 1 - 12 a; + 50 :»?- 72 y?- 21 x'+ 72 a^+ 50 'J^+ 12 x^-\- a^.

3. 16 a»-32 a' -4.0 a« + 88 a'-\- 49 a^-88 a'-AO a^+S2 a-|-16

^ Find the sixth roots of the following : y

4. x'^ + 6 x^Y + 15 x^Y + 20 xY + 15 xY + 6 a.Y" + y'l

5. a« - 12 a' + 60 a^ - 160 a^ + 240 a^ - 192 a + 64.

6. Find the fourth root of 209727.3616.

7. Find the sixth root of .009474296896.

THEORY OF EXPONENTS. 191

XX. THEORY OF EXPONENTS.

208. In the preceding chapters, an exponent has been
considered only as a positive integer.

Thus, if m is a positive integer,

a"' = axaxax ••-to m factors. (§ 6)

•

209. Let m and n be positive integers.

Then, a^ = a'X.axax • • • to m factors,

and a" = a X a X a X • • • to n factors.

Whence, a" x a^'^axaxax ••• to m + n factors.

That is, a"" v: a'' = a"'+". '^ * (1)

This proves the law stated in § 46 for all positive integral
V9,lues of the exponents.

Again, (a"*)" = O" x oT" X oT" X • • • to n factors

^m+m + m-\ — to n terms

That is, (a")" = «""'. (2)

This proves the first paragraph of the law stated in § 186
for all positive integi'al values of the exponents.

210. It is found convenient to employ exponents which
are not positive integers; and we proceed to define them,
and to prove the rules for their use.

It will be convenient to have all forms of exponents sub-
ject to the same laws in regard to multiplication, division,
etc. ; and we shall therefore find what meanings must be
attached to fractional, negative, and zero exponents in order
that equation (1), § 209, may hold for all values of m and n.

192 ALGEBRA.

211. Meaning of a Fractional Exponent.

1. Required the meaning of aK

If (1), § 209, is to hold for all values of m and n, we have

5 5. _5_ 5_L 5^ ■ 5 _

Hence, a^ is such an expression that its third power is a^

Then, a* must be the cube root of a^ ; or, a^ = Va^.

p
2. Required the meaning of a', where p and q are any

positive integers.

If (1), § 209, is to hold for all values of m and n, we have

■p. P P ^ ^ ?+?+?+ ...,o J terras "-xq

a^ xa^ X a^ X ••• to q factors = a" * « = a'' = a^.

p
Hence, a* is such an expression that its qth power is a*".

Then, a* must be the gth root of a^ ; or, a" = -^'a^.

Hence, in d fractional exponent, the numerator denotes a
power, and the denominator a root.

For example, a* = Va^; b^ = V^; x^ = V^; etc.

EXAMPLES. A

212. Express the following with radical signs :

1. a7. 3. 4a;*. 5. aM. 7. 6x^yK 9. abh^d"^.

2. bk 4. 9a&i o. ..^^i 8. 8aW. 10. Sx^y^zi

Express the following with fractional exponents : ^f ,\ ^(^

11. Va\ 13. Vm^ 15. 2A/n«. 17. </^'Vb\

12. </S. 14. Vb\ 16. 5-v^. 18. -v^^W,

19. 7v^V^. 20. Vo^VfiVc".

/ THEORY OF EXPONENTS. 193

^^ '-^ 213. Meaning of a Zero Exponent.

If (1), § 209, is to hold for all values of m and n, we have

1^ a"* X a^ = ^1"+" = a™.

, Whence, a" = — = 1.

a"

/ Hence, the zero power of any quantity is equal to 1.

214. Meaning of a Negative Exponent.

1. Required the meaning of a~^.

If (1), § 209, is to hold for all values of m and n, we have

a-3 X a^ = a-3+3 = a" = 1. (§ 213)

.^. 1

Whence,

.3

2. Required the meaning of a~% where s is a positive
integer or a positive fraction.

If (1), § 209, is to hold for all values of m and n, we have
a' xa' = a-+^ = a" = 1. (§ 213)

■rrr, 1

Whence, a~' = —

a'

For example, a~^ = ~ ; a~^ = — ; 3 x~^y~"^ = — - ; etc.

215. It follows from § 214 that

Any factor of the numerator of a fraction may he trans-
ferred to the denominator, or any factor of the denominator
to the numerator, if the sign of its exjwnent be changed.

Thus, ^=^ = ^ = ^,etc.

»*

EXAMPLES.

216. Write the following with positive exponents :

1. a?h-\

5. a~^x~*.

9. 6a;-V^-

2. m-hi\

6. 5 ma;"^.

10. Zx~~Hfz-i.

3. xY^.

7. 4.a-%-i.

11. 6 7>i-2?r^jA

4. 2ain-\

8. a"^&-V.

12. a-'6"'c"T«.

Transfer the literal factors from the denominators to the
numerators in the following :

13. 1. 15. _:^. 17. -^.

a6'

2/ ^2'

_4 .

14. ^. 16. — i 18. -^

19.

6 m^w-^

5p\

20.

3 5

3 a «a;^

8 6-^y^

ys 1 m~''x^ 4 & ^c

Transfer the literal factors from the numerators to the
denominators in the following :

oi Sx^ OQ «^*' on '^^'''y 07 8a"i'6-^

.6 9

22. ^. 24. ^^. 26. ^^I!^. 28. i^^M.

c* 6 y-l 9&-^?i~^

217. Since the definitions of fractional, negative, and zero

exponents were obtained on the supposition that equation

(1), § 209, was to hold universally, we have for all values of

m and n

a™ X a" = a"'+".

For example, o? x a~^ = a-~^ = a~^ ;

3 _2 3_2 J

a X Vc? = a X a^ = a^"*"^ = a^ ; etc.

THEORY OF EXPONENTS. 195

EXAMPLES.
Find the values of the following :

1. a^ X a"'. 4. m^ X m~K 7. 5x-^x4:X~*.

2. a^ X a~\ 5. 2 n^ x n~^. 8. 7«,^ x Vm.

3. a;-" X a;-*. 6. a x 3 a"'«\ 9. c^ x VcF^
10. «-i X — • 13. x-'y* X 4a;".v-l

w 5
11. 7^v^ X Vo'

14. m*n ^ X |^m~Vt~^.

_ 15. X a-^x K

12. SaVb' x2-\/b\ a-'x^

16. Multiply a + 2 a^ — 3 a^ by 2 — 4 a"^ — 6 a"i

a + 2 J - 3 J
2 -4a"^- 6a~^
2a + 4a3 - Ga^

- 4 a^ _ 8 a^ + 12

- 60"^ - 12 + 18a"^
2 a -20a3 +i8rt~^, ^ns.

Note. It must be carefully borne in mind, in examples like the
above, that the zero power of any quantity is equal to 1 (§ 213).

Multiply the following :

17. a^-\-a'b' + h^ by a' —bK

18. 4:X~^ -6 x~i + 9 by 2x~^ + 3.

19. 2 a-i - 7 - 3 a by 4 a-' + 5.

20. x~^ + 2 x~'^ + 4 a;"^ + 8 by a-"' — 2.
'\ 21. a;^ + x6?/7 4- y3 by a;3 — x^y^ + ?/^-

22. m — 2 m^?i"3 + m^?i"* by m^ri"^ — 2 m^n"^ 4- w"^.

196

ALGEBRA.

23. a-26-3 -I- a-^ft-* - a-'b-' by a-'b-- - a-''b-^ - a-^b-\

24. m-^+ 2 m~~hi-^+ 3 m'/w"^ by 2 m"' — 4 7i-'+ 6 m^n-'i

25. 2 a''6-2 + a^ - 4 a" V by 2 a^ - 6^ _ 4 ^-'^^

31 21 12 11 3

26. 3m^x^—4:mx^'-\-m^x by 6m^a;~3-)_8m~^a;~^ + 2m~fc

218. To prove — = a"*"" for all values ofm and n.
a"

By § 215,

= a*" X a-" = a"-", by (1), § 209.

For example,

= o"?-^ =

-•JO- 2 _ 1 3.

etc.

EXAMPLES.

Divide the following;

1. a* by a^.

2. X by a;3.

- 1 _t

3. m^ by 7» *.

4. a" 3 by (j?^

5. 6-2 by V6^.

6. 2a/^ by a;-l

7. n^ by ~

8. lOa-^b--^ by 5a^Z>-2,

9. 6a/^' by 2\/^.

10. Divide 2 a^ - 20 + 18 a"^ by a + 2 a^ - 3 a^.

2a^ -20 + 18a ^
2 a^ + 4 a^ - 6

a + 2a3 -.3a3

2 a "3 _ 4 a 3 _ 6 a'S ^ns.

-4a3 -14 + 18a~3
-4a^- 8 + 12ffl~^

- 6- 12a~3 + 18a"s

- 6-12a~^ + 18a"^

THEORY OF EXPONENTS. I97

Note. It is important to arrange the dividend, divisor, and each
remainder in the same order of powers of some common letter.

Divide the following :

11. a^ + &2 by a^ 4. &t. 12. a-i - 1 by a^ - 1.

13. .'K* - 2 + a;-* by or + 2 + x-\

14. a — 4a^ + 6 a^ — 4 a^ + 1 by a^ - 2 a* + 1.

15. x^ — Zxy^ -{-^ x^y^ —yhjx^ — yK

16. m~^ — 3 m~^ — 4 m~^ by m~^ + 2 m~^.

17. 9 ar^/ + 5 + a-" V"^ by 3 x-^ - x-^y-^ + x-^y-\

18. a~% — 5 am~^ + 4 a^wi"^ by a~^m? — a~^m — 2 a'^
J 9. a&~^ - 10 6^ + 9 a-^h by a^ + 2 &^ - 3 a'^fti

20. m^ — 2x^ + m'^a? by 7Ji~^a;* — 2 m'^o;^ + m~^aj«.

219. To prove {a"*y = a™" /o7' aZ/ values ofm and n.
We will consider three cases, in each of which m may
have any value, positive or negative, integral or fractional.

I. Let w be a positive integer.

The proof of (2), § 209, holds if n is a positive integer,
whatever the value of m.

P

II. Let n = -, where p and q are positive integers.

Then, by the definition of § 211,

(«"•)» = Viary = -y^(§ 219, I.) = a« .
III. Let n=—s, where s is a positive number.
Then, by the definition of § 214,

{cry = -1- = — (§ 219, I. or II.) = a-™.

^ ' {cry a"'^ ^

Therefore, the equation holds for aE values of m and n.

198 ALGEBRA.

For example, (a^)-^ = a^'^-' = cr^^ ;

(a~^)~3 = a~^^-i = a ;
( Va) 3 = (a^) » = a^^ 3 = (^3 ; etc.

EXAMPLES.
220. Find the values of the following:

^- («'r^- 7. (x-i)i 12. (^^)-t.

^- ^^^")'- 8. (</^)K 13. (J-

4. (m-*)l . _ 14. (a ^)««.

10. (^m)«. ' , ,

11. ^^^

_4

6. (a-i)"". V«/ 16- (a'" )™~"-

221. The value of a numerical quantity affected with a
fractional exponent may be found by first, if possible, ex-
tracting the root indicated by the denominator, and then
raising the result to the power indicated by the numerator.

1. Find the value of (- 8)i
We have,

(-8)t = [(- 8) 3]2 := (^378)2 = (_ 2)2 = 4, Ans.

EXAMPLES.

Find the values of the following:

2.

25l

6.

49-^.

10.

16-i

14.

32-^

3.

9l

7.

(-27)-i

11.

(-32)^.

15.

(-64)1

4.

8l

8.

4l

12.

64i

16.

(-243)1

5.

8ll

9.

343l

13.

(-125)1

17.

(-128)-7-

THEORY OF EXPONENTS. 199

222. To prove (aby = a"&" for any value ofn.
. I. Let « be a positive integer.

Then, (aby = ab x ab x ab x ••• to n factors

= (a X a X • • • to w f actors)(& x 6 x • • • to n factors)

II. Let w = -, where p and q are positive integers.
Then by § 219, [(aby]" = (aby = a^b^, by § 222, I.

And by § 222, 1., [«%«]« = (Jy(b^y = a^b^.

p p p

Therefore, [(«*)']' = [«'&']'. )

Taking the ^-th root of both members, we have

PIP
(aby = a»6*.

III. Let n = — 8, where s is any positive number.

Then, («&)-''= ^ = J- (§ 222, L or IL)=a-6-.
(aby a'b'

\/

'' MISCELLANEOUS EXAMPLES.

223. Square the following by the rule of § 78 or § 79 :

1. 2a^ + 36i 2. 5arY-2afy-^ 3. 3aV^-4a-V-

Extract the square roots of the following :
4. a-^bi. 5. 25mn-^xK 6. ^. 7. «'^"*

8. 4a^ + 4a*-19-10a"^ + 25a"i

9. 9a;"^-12a;-2 + 10a;"^ — 4x~* + l.

10. a^b-* - 8 ah-' + 10 a^-" + 24. ah-' + 9 aft"*.

'200 ALGEBRA.

Extract the cube roots of the following :

4

11. a*b-\ 12. xyh'K 13. ~ 27 m^rri 14. ^"'^

8y-i

15. 8a*-12aV^ + 6aV3-6-t

16. x~i 4- 6a;"^ + 3a;^ -28a;^ -9x2 + 540;^ _ 27£cl
I Simplify the following .

17.

20.

[ix-r-r^\

18.

a"+^ X a"~^

21.

19.

1 1 1

22.

X z-y i+y

(a'^+y H- a ^ ) * .

23. (2"+" - 2 X 2") X (2-2 x 2-"-2).

24. (a;"^)"''-"'-^('^Y-

25 (a;^-a^)''+(l + a;M)'

1 + a^a;-(a~^x^ — a^"^ — a^x^)

26. 4±X + _^I^. 28 x^^z£l-^Ll±£l
xi-yi xi^yi ' x'^-^-y-^ x'^ - y^

/ 3n _3n 3n

27. ^' —^ ^ 29 ^^ — 1 x" — 1

d'-a ' a;2-l a;2 + l

30. ^"' + y\«^-'-y~^_i.

a;-^ — y~^ x~^ + ^/-^
31 tt' + 2 6^ g^ — 2 a^lP +4 6^

RADICALS. 201

XXI. RADICALS.

224. A Radical is a root of an expression, indicated by a
radical sign ; as Va, or -yx + 1.

If the indicated root can be exactly obtained, the radical
is called a rational quantity ; if it cannot be exactly obtained,
it is called an irrational quantity, or surd.

225. The degree of a radical is denoted by the index of
the radical sign ; thus, V -c + 1 is of the third degree.

226. Most problems in radicals depend for their solution
on the following principle :

1 1 1
For any value of n, {abf = a" x h\ (§ 222)

That is, -Vab = -v/a x V&.

REDUCTION OF A RADICAL TO ITS SIMPLEST FORM.

227. A radical is said to be in its simplest form when the
expression under the radical sign is integral, is not a perfect
power of the degree denoted by any factor of the index of
the radical, and has no factor which is a perfect power of
the same degree as the radical.

228. Case I. When the expression under the radical sign
is a x)erfect power of the degree denoted by a factor of the index

1. Reduce VS to its simplest form.

We have, V8=\^ = 2^ (^ 211) = 2^ = v^, Ans.

EXAMPLES.
Reduce the following to their simplest forms :

2. \/36. 3. -^. 4. a/25. 5. -\M.

202 ALGEBRA.

6. ^. 9. ^2l6. 12. -v/121 a:%\ 15. -v/81 m«w^.

7. ^I6. 10. ^100. 13. ^125 a%«. 16. ^8^W.

8. v'l. 11. ^243. 14. ^32 a'^m^ 17. ^"27^^.

\ 229. Case II. When the expression under the radical sign
is integral, and has a factor which is a perfect power of the
same degree as the radical.

1. Keduce V54 to its simplest form.

We have, V54 = V27 x 2 = V27 x V^ (§ 226) = 3V2, Ans.

2. Eeduce VS a^b — 12 a^b"^ + 12 ab^ to its simplest form.

V3 a^b - 12 a262 + 12 a¥ = V(a2 _ 4 a6 + 4 &2)3 ab

= Va2 - 4 ai 4- 4 6 V3a^ = (a - 2 6) VS^, ^ns.

From the above examples, we derive the following rule :
Resolve the expression under the radical sign into ttvo fa^

tors, the second of which contains no factor luhich is a perfect

poiver of the same degree as the radical.

Extract the required root of the first factor, and prefix the

result to the indicated root of the second.

EXAMPLES.
Reduce the following to their simplest forms :

3. V28. 7. 3V98. 11. V375. 15. V192 mW

4. V99. 8. Vl50. 12. ^162. 16. a/'128 xyh'

5. V80. 9. 5-v^l08. 13. V128. 17. V64a*'6''.

6. -^/iO. 10. V243. 14. V242 a-U--. 18. V96a^6V.

19. Vl08a^« + 72 a^&^ 21. V{a'' -4.h-){a-2b).

20. -C/r35 ^f - 108W- 22. V5.r' + 30a;2 + 45a;.

23. V27 a^b - 36 a'b^ + 12 ab\

24. V(ar' -x- &){x' + 2 a; - 15).

RADICALS. 203

If the expression under the radical sign has a numerical
factor which cannot be readily factored by inspection, it is
convenient to resolve it into its prime factors.

25. Reduce Vl044 to its simplest form.

Vmi = \/23 X 3^ = v'28 X 33 X \/3^ = 2x3xv^ = 6\/9, Ans.

26. Eeduce V125 x 147 to its simplest form.

Vl25 X 147 = V53 x 3 x 7- = V6- x 7^ x V3 x 5 = 5 x 7 x VlE=35VTE, Ans.

Reduce the following to their simplest forms :

27. V864. 30. V125 x 135. 33. -v/4ll6.

28. V2625. 31. V98 x 336. 34. ■\/196 x 392.

29. V3528. 32. -v/Tl25. 35. ■v/40 x 45 x 48.

36. VToo^x 105 ab x 189 b^

230. Case III. When the expression under the radical
sign is a fraction.

In this case, the radical may be reduced to its simplest
form by multiplying both terms of the fraction by such an
expression as loill make the denominator a perfect power of the
same degree as the radiccd, and then proceeding as in § 229.

/"9~
1. Reduce A/x— ^ to its simplest form.

Multiplying both terms of the fraction by 2 a, we have

JX = >ZI« = J_9_x2a = JI^xV2^=A.^^,^„.
ySa^ > 16 a* \l6a4 \l6a* 4 a-

EXAMPLES.
Reduce the following to their simplest forms :

'• 4 3. J

I- ^-VS- -Vl

204 ALGEBRA.

13 - 4. 14. 4 18. Vf «'*'

6. A^- 10.

20 \9 V3 ^IS crd'

7. Jll. 11. ^5. 15. #. 19. JI^.

\32 \25 Vie \ 273^

9-4 i«-4 i^.ib- ^'-^^

4 ^8 \10.r5 \16^2

22.

5. J^±_^. 23. -^^J^

ar^ - 8 .^• + 8

.-i

^ 231. To Introduce the Coefficient of a Radical under the
Radical Sign.

The coefficient of a radical may be introduced under the
radical sign by raising it to the power denoted by the index.

1. Introduce the coefficient of 2 a V3 a^ under the radical
sign.

2 a\/Sofi = y/8a^\/3¥^ = y/S a^ x 3 x^ (§ 226) = \/Ua^, Ans.

Note. A rational quantity may be expressed in the form of a
radical by raising it to the power denoted by the index, and writing
the result under the corresponding radical sign.

EXAMPl ES.
Introduce under the radical signs the coefficients of :

2. 5V2. 5. 5^. 8. 4aV8a. 11. xY'-V^.

12. 3m=^A/2m.

13. 2a^'Ui\

3. 8V3.

6. 2^5.

9. Tic^Vear*.

4. 4-V/6.

7. 3V2.

10. Sah^5a\

14. (l + a)J^^^^- 16. ^^^:i^-J^±^.

^ ^^1 + a a + b^a-b

15. (.-i)V^+a. 17. ^Vi^Z!^,

232. Similar radicals are radicals which do not differ at
all, or differ only in their coefficients ; as 2 V ax^ and oVaaf.

RADICALS. 205

ADDITION AND SUBTRACTION OF RADICALS.

233. To add or subtract similar radicals (§ 232), add or
subtract their coefficients, and prefix the result to their
common radical part.

1. Required the sum of V20 and V45.

Reducing each radical to its simplest form (§ 229), we have
V20 + 745 = V4^5 + Vt) X 5 = 2 V5 + 3 VS = 5 V5, Ans.

2. Simplify ^ + ^|-^|

X2 \3 \8 \4 \£) Vlfi

16
= iV2 + i\/6-|V2=l\/6-|\/2, Ans.

E,ULE.

Reduce each radical to its simplest form.
Unite the similar radicals, and indicate the addition or sub-
tractio7i of those which are not similar.

EXAMPLES.
Simplify the following:

3. V75+Vi2. 5. V80-V180. 7. ^l92-^/3.

4. V98-V18. 6. -\/5l+Vl6. 8. </82-^162.
9. V27+V108-V48. 10. VTTS - Vll2 - Vii.

11. J-+J^. 12. JI+ J. 13. J+A^.
\2 \ 8 \27 \3 \9 \36

14. V5+V245-V32(). 16. a/5 - ^320 + -^72.

206 ALGEBRA.

18. b^VSaPb + ab V50 d'b' - aVl28 ab^

19. m^ v^32m2 + m^l08 vv" + ^50Ch^.

20. V50 a* - 75 a^o; - V32 a V - 48 o.-^.

23. ■v^--v/24--v^+V^75.

24. ■^v/243-v'l8--v^68.

25. V32-V72+V125+ a/162 -V500.

26. ar* Vl50« + V96 x^ - V54 x^ - x V24 a;^.

27. V63a2^ + &VlM^-V40aP-aA^252&.

"■xi+ViWl-4

'' 30. V80 ar^ + 40 0^ + 5 .'K + V45 a;^ _ 60 x-^ + 20 ic.
31. 2Vl2a.'2 + 60a:2/ + 75/-V48a,-2-72a;2/ + 27i/2.

32. J«T6_J^+-^^V^^-::^^.

^a — & ^a + 6 a^ — ¥

TO REDUCE RADICALS OF DIFFERENT DEGREES TO
EQUIVALENT RADICALS OF THE SAME DEGREE.

234. 1. Eeduce V2, v^, and -x/S to equivalent radi-
cals of the same degree.

By § 211, V2 = 2^ = 2t'^ = v^26 = v^ ;

^ = 5* = 5T2 = ^/p = v^l25.

RADICALS. 207

We then have the following rule :

Express the radicals with fractional exjwnents, and reduce
these exponents to a common denominator.

Note. The relative magnitude of radicals may .be determined by
reducing them, if necessary, to i-adicals of the same degree.

Thus, in Ex. 1, Vi25 is greater than v8], and v81 than v64.
Hence, Vs is gi'eater than \/3, and \/3 than V2.

EXAMPLES.
Eeduce to equivalent radicals of the same degree :

2. V3 and VS. 7. V^, '^y^i and Vi^.

3. V2 and ^. 8. ^/Wa, </2h, and -s/^.

4. a/o^ and ^a^. 9. ^2, -Vs, and ^3.

5. V2 and V^. 10. </l^^ and -^T+^.

6. Vi and V6. H. VaH- 6 and -y/a — b.

12. Which is the greater, a/2 or "v^ ?

13. Which is the greater, Vll or VS ?

14. Which is the greater, VlO or Vi ?

15. Arrange in order of magnitude, Vl4, V6, and V1T5.

16. Arrange in order of magnitude, a/3, a/IS, and a/2o3,

17. Arrange in order of magnitude, a/3, a/5, and a/7.

MULTIPLICATION OF RADICALS.
235. 1. Multiply a/6 by ^/15.

V6 X VT5 = V6"x 15 (§ 226) = V2 x 3 x 3 x 6
= \/32 X V2ir5 = 3a/I0, ^ns.

208

^■^ 'IdX^^

ALGEBIU. . ^

2. Multiply V2a by -^4 a'
Reducing to equivalent radicals of the same degree,
/2a X VV^^-i:! of X (4 a^)

:(2a)6 x(4a2)6 = V(2a)3x V(4a2)2

= \/23a3 X 2%* =: V^6a6 X V2 a = 2 a V2 a, ^ms.

3. Multiply Vl'O by ^5.

We have, \/20 == 20^ = 20« = V2P = V(22 x 5)3 = V26 x bK
Whence, V20 x \/5 = \/26 x 5^ x 5 = v''26 x \/5* = 2 x 5^ ^ 2 x 65-
= 2 x-v^52 = 2v^, ^«s.

.From the above examples, we have the following rule :
Reduce the radicals, if necessary, to equivalent radicals of

'the sarae degree.

Multiply together the expressions under the radical signs,

and write the result under the common radical sign.
-The result should be reduced to its simplest form.

EXAMPLES.

Multiply the following ;

4. V3 and V48.

5. V6a^ and V36a.

6. Vl4 and Vl8.

7. Vl5 and VSO.

8. V44 and V275.

9. VSOab and V70 ac.

10. ^36 and "v^.

11. -s/IE and \/63.

12. -v/90 and Vi32.

13.

/21 , /35

14. ^ and A/i35.

15. -v/48 and V56.

16. V3a; and V27^.

17. V6 and a/12.

18. VSax and VJte.

19. V5 and v'i25.

20. -v/r^ and ^^/8¥i.

21. a/45 and a/9.

22. Vl2 and ^^3.

23. a/54 and a/72.

RADICALS. 209

24. ^ and ^. ^7. VB, ^2, and ^8.

,^ ,^ / 28. V2, ^4, and -^3.

25. J^ and ^y^.

26. Vab, -^hc, and "v^ca. ' ^19 ^'8

30. Multiply 2V3+3V2 by 3 V3 - V2.

2V3 + 3V2
3V3- V^

I8 + 9V6

-2a/6-6
18 + 7V6-e = 12 + 7V6, ^ns.

Note. It should be remembered that to multiply a radical of
the second degree by itself simply removes the radical sign; thus,
V3 X V3 = 3.

31. Multiply 3Va^ + l +4 a; by 2^x^ + 1— x.
3\/x2 + 1 + 4x
2v'x2+ 1 - «

6(a;2 + i) + 8xVx-^ + l

-3xVx2 + 1 -4x2

6x2 + 6 + SxVx- + 1-4x2 = 2x2 + 6 + 6xVx2+ 1, Ar

Multiply the following :

32. 5 - 2 V3 and 4 + 3 V3.

33. 2Vx+-3V2 and Q^-^.

34. 7 V2 - 4V5 and 3V2 - 8V5.

35. 6Va + llV& and 9Va-5V&.
'i 36. 5^/4 + 3^/5 and 6^/2 + 7-v^.

37. Va + 2V&-3V^ and V^-2V5-3Vc.

38. 4Vic-|'l -SVcc-l and 3 Va; + 1 - 2 V^^^.

210 ALGEBRA.

39. V6 - V3 - V5 and VO + V3 -f Vs.

40. 9^-8^1 and 3^ + 104

41. 5 Va + 3 V5 + 4V7 and 5 V3 + 3 VS - 4V7.

42. 3 V2 - 4V5 - 2 V7 and 6 V2 - 8 VS + 4 V7.

43. oV8 + 6Vl2-2V20 and 7V2 -3V3 + 4V5.

«. eVi-a-vl+ioVl and 2^1+3^1-54

Expand the following by the rules of §§ 78, 79, or 80:

45. (3Vo+4)2. 48. (7Vl0 + 5V7)l

-i«. ^^-2^^)". 49. (V2a+V3a-4y^.

47. (6V2-4V6)2. 50. (3V^4^ - 2 V^T^/.

51. (3V2 + 7)(3V2-7).

C52. (6V3+4V5)(6V3-4V5).

53. (2V^Tl + 5V^)(2V^Tl-5v^).

54. (Va + & + Va - 6)(Va -f- 6 - Va - &).

55. (3V2a - 5 + 4V4a- 3)(3V2a - 5- 4V4a - 3).

DIVISION OF RADICALS.
236. By § 226, v^= Va x V6.

Wlience, '^=-Vb.

A'a

We then have the following rule :

Beduce the radicals, if necessary, to equivalent radicals of
the same degree.

Divide the expression under the radical sign in the dividend
by the expression vjider the radical sign in the divisor, and
write the result under the common radical sign.

RADICALS. 211

EXAMPLES.

1. Divide Vl5 by VO.

Reducing to equivalent radicals of the same degree, we have

2. Divide VlO by ^/40.

We have, VlO = 102 = I0<r = VIO^ ^ V(2 x 5)3.

Whence, = A/ • = v5^ = 5*f = 5^ = v5, Ans.

^iO ^ 23 X 5

Divide the following:

3. V84 by V7. 5. V56 by V32. 7. ^5 by ^135.

4. Vl2 by Vl5. 6. a/IG2 by ^2. 8. a/63 by ^35.
9. A/8i by A^'9. 19. V6a^ by ^/Oa^.

10. A/26a by -^39a^. ^ /^ .rr

^ 20. A^ by V/-i.

11. -^192 by </3. ^^ ^^^

12. a/2 by ^8. 21. ^/i ^y . /? .

13. V5 a& by V12o i'c.

14. V 28 X by V42 cc. ^ j v i,
ic /^i T. /28

16. V2Mby V2jr. """ \ 5 -" \5

21 , i28 ^^ ^-^ ^y ^1-^-

17. -?/3 w by \/7 m«. 25. -v/12 a' by V8 a'.

3/12 , a/lT

18. ^»/12by^. 26. Jl by 41.

^ \20 \8 -^ ^16

212 ALGEBRA.

INVOLUTION OF RADICALS.

237. 1. Raise -\/l2 to the third power.

( Vl2f = (12^)3 = 12« (§ 219) = 12^ = VT2 = 2\/3, Ans-

2. Raise V2 to the fourth power.

( v^)4 = (25)4 = 2* = v^2* = v^, Ans.

We then haye the following rule :

If possible, divide the index of the radical by the exponent
of the required p)0iver; otherwise, raise the expression under
the radical sign to the required power.

EXAMPLES.

Find the values of the following :
3. (^3)*. 7. {2V^y. 11. (\/l28)3.

10

4. {-VSf. 8. {-VTay. 12. (5mV96<)2.

5. {V^^+yy. 9. {-^jW^y. 13. (a/3 a -2)3.

6. {3-smf. 10. {-Vsy. 14. ('v'48^)^

EVOLUTION OF RADICALS.

238. 1. Extract the cube root of v27 a?.

v'(\/27x^0 = (^^C3x)3)^- L(3a;)^]5=(3x)'S"= v^, ^»s.
2. Extract the fifth root of -s/Q.

\/( \/8) = (63 )* = 6T^ = \/6, ^«s.

We then have the following rule :

If possible, extract the required root of the expression under
the radical sign; otherwise, multiply the index of the radical
by the index of the required root.

Note. If the radical has a coefficient which is not a perfect power
of the degree denoted by the index of tlie required root, it should be
introduced under the radical sign (§ 231) before applying the rule.

Thxjs, ^/W^ = -s/V32 = V2.

RADICALS. 213

EXAMPLES.
Find the values of the following:

3. V(a/49). 7. V(16^). 11. -^/(2a</2^).

4. a/(V10). 8. </(-y/2E). 12. </ (27 a^ \/5x).

5. </(</32^'). 9. V(a/x-2-6.t+9). 13. </(</T25).

6. V(V81a;Y). 10. </i3</2xy'). 14. V(3a2V9a).

TO REDUCE A FRACTION HAVING AN IRRATIONAL
DENOMINATOR TO AN EQUIVALENT FRACTION WHOSE
DENOMINATOR IS RATIONAL.

239. Case I. mien the denominator is a monomial.

The reduction may be effected by multiplying both terms
of the fraction by a radical of the same degree as the de-
nominator, having under its radical sign an expression
which will make the denominator of the resulting fraction
rational.

5
1. Reduce '' to an equivalent fraction having a

V3 a^
rational denominator.

■ Multiplying both terms of the fraction by Vda, we have
5 5v^9a 5\/9a 5\/9a

f^' v^^v'Oa Wfa^ 3a

-, Ans.

EXAMPLES.

Reduce each of the following to an equivalent fraction
having a rational denominator :

2. A. 4. -^. 6. ^4^. 8.

V6 V25 V4 a%

V'7a63 ^3cc2 a/27 ^l<6:i?yz^

214 ALGEBliA.

240. Case II. When the denominator is a binomial con-
taining radicals of the second degree only.

1. Reduce — to an equivalent fraction having a

5+ V3
rational denominator.

Multiplying both terms of the fraction by 6 — VS, we have

6-v/8_ (5-V3)2
6 + VS ~ (5 + V3)(6 - V3)

^26-10V3 + 3 79 g ^28-10V3^14-lV3 ^,^^^
26-3 ^S8 ' ' 22 11

2. Keduce — to an equivalent fraction hav-

2V^ — 3Vaj — 1
ing a rational denominator.

Multiplying both terms of the fraction by 2-\/x + 3v/« — 1,

2Vx-Sy/x~- 1 (2 V« - 3\/«^^)(2 Vic + 3Vx - 1)

_6!g-t-5v^Va:-l -6(a;-l)_6+ 5\/x- ~~x ^
" 4a;-9fa;-l) 9-5x '

AYe then have the following rule :

Multiply both terms of the fraction by the denominator icith
the sign betiveen its terms changed.

EXAMPLES.

Reduce each of the following to an equivalent fraction
having a rational denominator :

8 6 ^ V^-Vy g 3V3-2V5

S+v'S Vx + Vy ' 5V3 + 4V5

4 5 7 W2+V6 jQ V^^^ + 1

* 2V3-4* * 3V2-V6* ' V^^=^ + 2*

- Va + 5 « 3V5 + 2V2 ,, Va-6+Va

O. — ;= • O. ;= —• 11. — ; 7=*

Va-b 3V5-2V2 ^a-b-^/a

EADICALS- 215

12 2^+V^ — 4ar^ ^e Va; — y + Vo; + y^

2 a; — Vl — 405* Va; — y — Va; + 2/

13 <* — Va^ — 6'^_ jg Va^ — ar — Va^ + x^

a + Va^ — b^ Va^ — a.*^ + Va^ + ar^

14 Vl +a —VI — a^ •,» 4 Va; — 1 + Va; + 1

Vl+a+Vl— a SVa; — 1 — 2Va; + l

241. The approximate value of a fraction whose denomi-
nator is irrational may be conveniently found by reducing
it to an equivalent fraction with a rational denominator.

1. Find the approximate value of to three places

of decimals. ^ "" ^^

^^1 _ 2+V2 -2 + V^_2 + 1.4111::^ 1707 ...,^ns.

2 -a/2 (2-V2)(2+V2) 4-2

EXAMPLES.

Find the approximate value of each of the following to
three decimal places :

o 3 5. 2 g 3V2-V6

VIO i^ 3V2+V6

3. _JL_. 6 5 g 3V3 + 2V5

V6-2 ' V3-2V2 ' 3V3-2V5*

4 _A__. 7. V5+V2. 10 ^-^^-

3 + 2V5 * V5-V2' ' 2V6 + 5V^3*

PROPERTIES OF QUADRATIC SURDS.

242. A duadratic Surd is the indicated square root of an
imperfect square ; as V3.

243. A quadratic surd cannot be equal to the sum of a
rational expression and a quadratic surd.

216 ALGEBRA.

For, if possible, let Va = b + Vc.

Squaring botli members, a = 6^ + 2 6 Vc + c.

Transposing, 2 bVc = a — b- — c.

Dividing by 2&, Vc = ~ ^ ~ —

2 b

We then have a quadratic surd equal to a rational ex
pression, which is impossible.

Hence, Va cannot be equal to 6 + Vc.

244. Ifa-\- V& = c + Vd, f/ien a = c, and V& = V^.

If a is not equal to c, let a = c + x.

Substituting this value in the given equation, we have

c+ x-\- Vb = c + Vd,
or, X + V6 = Vd.

But this is impossible by § 243.
Hence, a = c, and therefore V& = Vd.

245. ijr Va+V& = V^ + Vy, then -\/a — Vb= Vx — Vy.
Squaring both members of the given equation, we have

a + Vft = ar + 2 V^ + y.
Whence by § 244, a = x + y, (1)

and V& = 2V^. (2)

Subtracting (2) from (1),

a — Vft = X — 2V^ + y.
Extracting the square root of both members,

"^a — Vft = V» — A''^.

RADICALS. 217

246. Square Root of a Binomial Surd.

The preceding principles may be used to find the square
root of a binomial surd whose first term is rational.

Example. Kequired the square root of 13 — Vl60.

Assume, Vl3 - VIBU = Vx - Vy. {I)

Then by § 245, Vl3 + V\m =\^ + Vy. (2)

Multiplying (1) by (2), \/l69 - 160 = x - y.

Or, x-y = 3. (3)

Squaring (1), 13 - Vl60 = x- 2Vxy + y.

Whence by § 244, x + y = 13. (4)

Adding (3) and (4), 2x = 16, or x = 8.

Subtracting (3) from (4), 2 ?/ = 10, or y = b.

Substituting in (1), Vl3-\/l65 = V8-\/5 = 2\/2-\/5, Ans.

247. Examples like that of § 246 may be solved by in-
spection by putting the given expression into the form of a
perfect trinomial square (§ 96), as follows :

Reduce the surd term so that its coefficient may be 2.

Separate the rational term into tivo jjarts ivhose product
shall he the expression under the radical sign of the surd term.

Extract the square root of each part, and connect the results
by the sign of the surd term (§ 97).

1. Extract the square root of 8 + V48.

We have, Vs + ViS = Vs + 2V'l2.

We then separate 8 into two parts whose product is 12.
The parts are 6 and 2.

Whence, V8+\/48 = Ve + 2 \/12 + 2 = \/6 + V2, Ans.

2. Extract the square root of 22 — 3V32.

We have, V22 - 3\/32 = V22 - \/9 x 8 xl = V22 - 2\/72.
We then separate 22 into two parts whose product is 72.
The parts are 18 and 4.

Whence, V22 - 3 V32 = VIS - Vi = 3 \/2 - 2, Ans.

218

ALGEBRA.

EXAMPLES.

Extract the square roots of the following:

3. 11+2V28. 9. 12 -VIM 15. 56 + 5V48.

10. 11+V120. 16. 35-12V6.

11. 26 + 2V160. 17. 37-V640.
'12. 2O-6V1T. 18. 35-20V3.

13. 46-3V20. 19. 85 + 5V120.

14. 35 + lOVlO. 20. 75-3V96.

4. 17-2V72.

5. 49f2V48.

6. 19 + 4V2T

7. 28-8V6.

8. 30-2V56

■^r^ %.

21. 2 a; + 2^^^^.

22. a -2 ^ab - b\

IMAGINARY NUMBERS.

248. An even root of a negative number is impossible;
for no number when raised to an even power can produce
a negative result (§ 186).

An Imaginary Number is an indicated even root of a
' negative number; as V— 4, or s/ — a?.

In contradistinction, all other numbers, rational or irra-
tional, are called real numbers.

249. Every imaginary square root can be expressed as
the product of a real number by V— 1.

Thus, V— a^^Va^ x(— l) = Va^ X V— l = aV— 1;
V— 5 =V5 x(— 1) = V5 X V— 1; etc.

250. To find the positive integral powers of V— 1.
By § 189, V— 1 signifies an expression whose square is
equal to — 1.
That is, (V^' = - 1.

RADICALS.

219

Then,

(V^)^=(V::^)^x (V^)^=(-l) X (-1) =1 ;

(V^/=(V^)^x V-l = 1 xV^-V^I; etc.

Thus, the first four positive integral powers of V— 1 are
V— 1, —1, — V— 1, and 1; and for higher powers these
terms recur in the same order.

251. Addition and Subtraction of Imaginary Numbers.

Imaginary numbers may be added or subtracted in the
same manner as other radicals. (See § 233.)

Add

1. Add ^/-i and v-36.

y/Zri + x/ZTsG = 2 V"^ + 6 V^l (§ 249) = 8V'^^, Ans.

EXAMPLES.

Simplify the following ■

2. V^Il6 + V^25.

.1 + V^:49-a/-64.

3. V^+V^^.

5. Vi:^_V-(a-6)2.

6. v^r^'+V^2+V^=2^

8. V^l36-V^^100 + V^^ST.

9. V-a^-V-4a2-V^9^.

10. V-20 + V^^A5-V^^W.

11. V-l-2a;-a^-V^^4^.

252. Multiplication of Imaginary Numbers.

The product of two or more imaginary square roots may
be found by aid of the principles of §§ 249 and 250.

1. Multiply V^^ by a/"^.

v^ 2 X V^^ = V2 \/^nr X \/3 V^ (§ 249)

= y/2V3 ( V^)2 = V6 (- 1) (§ 250) = - V6 , Ans.

220

ALGEBRA.

2. Multiply together V^^, V-16, and V^=^.

v^ X V- 16 X v^2E = sv^ X 4\/^n; x 5 v^^

= 60(^/^1)3 = 60(- V^^)(§ 250)

3. Multiply 2 V^=^ + V^^ by V^ - 3 V^.

V^r2_3V^5

2(-2)+ V2 \/5(\/^=T)2

- 6V2 VI ( V^^)2 -3C- 5)

-4-5\/l0(- 1)

+ 15 = 11 + 5VT0, Ans.

Note. It should be remembered that to multiply an imaginary
square root by itself simply removes the radical sign ; thus,

V^ X V^^ = - 2.

EXAMPLES.
Multiply the following :

4. V^Tli) by V"=^. 8. -V^27by V"=12.

5. V^r9^2 by _ ^/3i6"(?. 9. _Vi:72by -V^^50.

6. V^Te by V^^IO. 10. 2-5V^ by 3+4V^^.

7. _V^r^ by -V^=T. 11. 8 + V^=2by 7-5V^.

12. 4V^=r5-7V^r3 by 2V^^-V^^.

13. 2V^=^ + 3V^^ by 4V=^-6V^=^.

14. V— a^, V— ?/^ and V— 2^.

15. V^ -V^^lS, and V^^32:

16. V^T0"+5V^^ by 3V^^ +

17. V^=^ + -

3 by V^8'-V-12.

RADICALS.

221

18. V-1, V-'S6, V-64, and V-100.

19. V^ V^^, -V^ and V^f.

Expand the following by the rules of §§ 78, 79, or 80:
20. (l+V^^^. 23. (2V^^ - 3 V=^)'.

7 21. (5V^^ + 2V^^y. 24. (x+V^)(x-V^).
22. (4-V^^)=^. 25. (5 + 6V^(5-6V^).

26. (3 V^ + 2 V^) (3V^a - 2 V^.

27. (7V^1^ + 4V^^) (7 V^=^ - 4V^.

28. (V^2 + 2 V^' + (V^^ - 2-\A=l)2.

Reduce each of the following to an equivalent fraction
haviiig a rational denominator :

29.
30.

l+V-3

1+V^

31.

32.

2+4V-6

5V-2-4V^=^

1-V-l 3V-5 + 2^^:^3

Expand the following by the rule of § 188 :
33. (1+V'^^)='. 34. {2-\'^^y.

253. Division of Imaginary Numbers.

1. Divide V^=^ by V^^.

We have,

V^^l2 VY2^

'\2

- 3 VSV- 1 V3

2. Divide VlO by V^^.
VlO ^-a/T0(-1).

= Vi = 2, .4ns.

IOC-

l)^(§250) = -v'uV-l = -v/^,

V2>/^ ^ns.

222 ALGEBRA.

EXAMPLES.

Divide the following :

3. V^25 by V^^. 8. -Va by V^=T^.

4. V^^32 by V^^. 9. </^^5 by -y/^^.

5. V42 by V^^. 10. -a/^^18 by ^/^^.

6. V63 by -V^7. 11. ■\/- 108 by a/^^.

7. V=^byV^=^. 12. -a/^=1o by ~-v/^.

SOLUTION OF EQUATIONS CONTAINING RADICALS.

254. 1. Solve the equation Va.-^ — 5 — x = — l.

Transposing — x, y/ofi — 5 = x — 1.

Squaring both members, x"^ — b = x^ — 2x + \.
Transposing and uniting terras, 2 x = 6.
"Whence, x = 3, Ans.

2. Solve the equation V2 a; + 14 +V2a? + 35 = 7.

Transposing \/2 x + 14, v'2 x + 35 = 7 — V'2x+ 14.

Squaring both members, 2x + 35 = 49 — 14V2x+ 14 + 2x + 14,

Transposing and uniting terms,

14\/2x + 14 = 28.

Or, V2X + U = 2.

Squaring both members, 2 x + 14 = 4.

2x = -10.
Whence, x = — 5, Ans.

From the above examples, we derive the following rule :
Transpose the terms of the equation so that a radical term

may stand alone in one member; then raise both members to a

power of the same degree as the radical.

If radical terms still remain, repeat the operation.

Note. The equation should be simplified as much as possible
before performing the involution.

RADICALS.

223

EXAMPLES.

Solve the following :

3. V3.r-5-2 = 0.

7. Vx + 4.+Vx = 3.

4. ■</Qx + 9 + 8 = 5.

8. V8ar^-12a;=' + l = 2;

5. V9a;2 + 5-3aj = l.

6. V^ - Va;-12 = 2.
3

9. V5a; + 10-V5^ = 2.'
10. Va; + 11 + ViB + 6 = 5.

11.

V6 — X.

V 3 — X VG — X
12. V2a;-l+V2a; + 4 = 5. 15. Vx+5'-}-Vx^^=2Vx.

13. V^ + Va; + 4 =

14. Va;-6+V^ =

-\/x
3

16.

17.

2V^-3^4V^-4
SV^+2 6VX+1

^x — 6 V3 a; + 1 — V3^

18. Va^-5a;-2+Var'4-3a; + 6 = 4.
10

= 4.

l9/ V^ — Va; — 5 =

21 V^ + « — Va; — g _ 2

Va; — 5 Vx + a + Va; — a

20. Vx'+T5-Vx + S = 2Vx. 22. V^+^+Va;-a = 26.

23. Va; + 2a-V^3a=-

Va; — 3 a

24. Va; + 6+Va; + 9 = V4a; + 29.

25. Va; + a+Va; + 4a = 2 Va; + 2 a.

26. Vc^~+ a& + Vca; — ab = VT

2a6.

27. V^ + '^a — Va a; + a;- = Va.

28. V(a; + 2 a V4 a; + 3 a^) = V^ - 2 a.

29. 2 (a; + a)(a; + Var' - a') = al

30. Va; + 1 +Va; + 5=Va; + 2+Va; + 3.

31. Va — a; + V6 — a; = Va + 3 6 — 4 a;.

224 ALGEBRA.

XXII. QUADRATIC EQUATIONS.

255. A Quadratic Equation is an equation of the second
degree (§ 158).

A Pure Quadratic Equation is a quadratic equation in-
volving only tlie square of the unknown quantity ; as
2 a^ = 5.

An Affected Quadratic Equation is a quadratic equation
involving both the square and the first power of the un-
known quantity ; as 2 a;^ — 3 a; — 5 = 0.

PURE QUADRATIC EQUATIONS.

256. A pure quadratic equation may be solved by reduc-
ing it, if necessary,. to the form x^= a, and then extracting
the square root of both members.

5^.2

1. Solve the equation 3 ar -f 7 = f- 35.

4

Clearing of fractions, 12 x- + 28 = 5 a;^ + 140.

Transposing and uniting terms, 7 x^ = 112.

Or, x2 = 16.

Extracting the square root of both members, we have

a; = ± 4, Ans.

Note 1. The sign ± is placed before the result, because the squai'e
root of a number is either positive or negative (§ 192).

2. Solve the equation 7 x^ — 5 = ox^ — 13.
Transposing and uniting terms, 2 x^ = — 8.
Or, x2 = - 4.
Whence, x = ±V- 4: = ±2V^^, Ans.

Note 2. In Ex. 2 the values of x are imaginary (§ 248) ; it is
impossible to find anj' real values of x which will satisfy the given
equation.

QUADRATIC EQUATIONS. 225

EXAMPLES.
Solve the following equations :

3. 3a;2-26 = 9a;--80. 5. 3(x + l) - x(x-l) =4.x,

4 _L_J1 = 5 g ^ , 3^^ , 12

* 3 x'2 9 a^ 8 ■ 3 a; 12 a;

7. (2 a; - 3) (a; + 7) + (2 a; + 3) {x - 7) = 58.

8. 5(x-\- a)(x — a) -]-4:ax= (x + 2 ay.

9. (3a.- + 2) (4a; - 5) - (5a; + 3) (6a; - 5) + 45 = 0.

.(. 2ar^ + 4 3a;2.-7 11 ,„ > /^l"^ 5a-

10. = — • 12. .-c+Va' + a;-^

5 3 15 Va2 + a;2

11. VIO + x- VlO - a; = 2. 13. V2a; + 8 + 2V« + 5 = 2.

,. x^-^x-\-l af-x + 1

1*. ij — j — = 0.

a; — 1 a; + l

15 3a;''-2 5a;^ + 3 4a;^-4^Q

5 10 25 '

16 9^-2^3a;^ 6a;^ + l

6 2 9a;2 + 7"

17 g?4-4 a;-4^10 ^g ^i" -Sx" + 4: _ x' ~3

x-4 x + i 3 3a;^+2a;2_4 Sx'-\-2

19. a;Va;2 + 12 + a;V^M^ = 3.

2Q a; + g a; — a _ -i , a^ + 2b^

X — b X -{- b a? — b^

21. Vl+a' + a^-Vl-a; + a;2 = V6.

22 ^"f" '^ I ^ — CI' _a + b a — b

X — a X -{- a a — b a + 6

(First add the fractions in the first member ; then the fractions in
the second member.)

226 ALGEBRA.

AFFECTED QUADRATIC EQUATIONS.

257. An affected quadratic equation may be solved by
reducing it, if necessary, to the form x- -\-px = q.

We then add to both members such an expression as will
make the first member a perfect trinomial square (§ 96) ;
an operation which is termed completing the square.

258. First Method of Completing the Square.

Example. Solve the equation a;- + 3 ic = 4.

A trinomial is a perfect square when its first and third
terms are perfect squares and positive, and the second
term plus (or minus) twice the product of their square
roots (§ 96).

Then, the square root of the third terra is equal to the
second term divided by twice the square root of the first.

Hence, the square root of the expression which must be

q „ q

added to a^ -\- 3x to make it a perfect square is — , or -•

Q ^ X Li

Adding to both members the square of -, we have

Extracting the square root of both members (§ 97),

a; + ? = ± -. (See Note 1, § 256.)

3 3,5 35

2- "" = -2 + 2' °' -2-2

Whence, a; = 1 or — 4, Arts.

From the above example, we derive the following rule :

Reduce the equation to the form x^ +px=q.

Complete the square by adding to both members the square
of half the coefficieyit of x.

Extract the square root of both members, and solve the sim-
ple equations thus formed.

QUADRATIC EQUATIONS. 227

259. 1. Solve the equation 3x^ — 8a; = — 4

Dividing by 3, x^--^ = -^;

which is in the form x'^ + px = q.

Adding to both members the square of f, we have

"^ 3 +I3J - 3+9'-9

4 2
Extracting the square root, x — = ± -•

o o

4 2 2

Whence, a; = -±- = 2or-, J.?is.

3 3 3

If the coefficient of x^ is negative, the sign of each term
must be changed.

2. Solve the equation — 9 x^ — 21 a? = 10.

Dividing by - 9, x2 ^ — = - — .

° "^ ' 3 9

Adding to both members the square of |, we have
^2 , 7x , /7\2 10 , 49_ 9

"^ +T^Uj ""¥+36-36

7 3

Extracting the square root, x + - = ± -•

6 6

7 3 2 5

Whence, x = — ±- = — or — , Ans.

6 6 3 3

EXAMPLES.
Solve the following equations :

Z. x' + 6x = 7. 10. 2a^ + lla; = -5.

4. x^-4:X = 32. 11. 2ay' + 9x-5 = 0.

5. a;2 + 11 x = - 18. 12. r)x- + S = 22x

6. aj^-lS.'c^-SO. 13. 20 -27 a; = -Gar'.

7. ar^ + a; = 30. 14. 7-10a;-8a;2 = 0.

8. 3^--7a; = -2. 15. 12 + 16a;- 3a;2 = 0.

9. 4a,'2-3a;=7. 16. 6ar' + 4 = - 11a;.

228 ALGEBRA.

260. If the coefficient of a;' is a perfect square, it is con-
venient to complete tlie square directly by the principle
stated in § 258 ; that is, by addyig to both members the square
of the quotient obtained by dividing the coefficient of x by twice
the square root of the coefficient of a^.

1. Solve the equation 9 o;^ — 5 a? = 4.

Dividing 5 by twice the square root of 9, the quotient is f .
Adding to both members the square of f, we have

n o ^ , /5\2 . , 25 169
9x2 — 5x+- =4 + --=: — — •
\Ql 36 36

5 13
Extracting the square root, 3 x — = ± - —

6 6

Transposing, 3a; = ^l:-;- = 3or— -•

DO o

4
Whence, a; = 1 or — , Ans.

If the coefficient of a? is not a perfect square, it may be
made so by multiplication.

2. Solve the equation 8 a^ — 15 a; = 2.

Multiplying each term by 2, 16 x^ — 80 a: = 4.

Dividing 30 by twice the square root of 16, the quotient is -\*'-, or ^.

Adding to both members the square of J/-, we have

1^ 2 on , /15\2 . , 225 289
16x2-30x+^-) =4+-==-.

15 17

Extracting the square root, 4 a; = ± -7-

4 4

Transposing, 4x = — ± — = 8or — -•

"Whence, x = 2 or - -, Ans.

o

Note. If the coefficient of x- is negative, the sign of each term
must be changed.

QUADRATIC EQUATIONS. 229

EXAMPLES.
Solve the following equations :

3. 4a^ + 7.T = 2. 10. 49a^-7a; = 12.

4. 16a^ + 32a; = -15. 11. 2rjx^ + 25x+ 6 = 0.

5. 9x2-llx = -2. 12. 12a^ + 8a; = -l.

6. 8a;2 + 2aj = 3. 13. 32a^ + l = - 12a;.

7. 5x' + 16x = -3. 14. 28+5a;-3a^ = 0.

8. 36 x-2 - 36 rK = - 5. 15. a; + l = 20a.-2.

9. 64a-2 + 48a;=7. 16. 4 + 3a;- 27a.-2=0.

261. Second Method of Completing the Square.

Every affected quadratic can be reduced to the form

acc^ -\-bx = c.
Multiplying both members by 4 a, we have

4 a^a^ + 4 abx = 4 ac.
Completing the square by adding to both members the
square of (§ 260), or b, we obtain

4 dV 4- 4 abx + 6^ = 6^ + 4 ac.
Extracting the square root,

2 aa; 4- 6 = ± V6^ + 4 ac.

Transposing, 2ax = — b± V6^ + 4 ac.

Whence, ^^-&±VPT4^

2a
From the above example, we derive the following rule :

Reduce the equation to the form aa? -{-bx = c.

Multiply both members by four times the coefficient of x?,
and add to each the square of the coefficient of x in the given
equation.

230 ALGEBRA.

Extract the square root of both members, and solve the
simple equation thus formed.

The advantage of this method over the preceding is in
avoiding fractions in completing the square.

262. 1. Solve the equation 2x^ — 1 x = — ^.

Multiplying both members by 4 x 2, or 8,
16x2-56x = -24.
Adding to both members the square of 7, we have

16x2- 56x + 72 = -24 + 49 = 25.
Extracting the square root, 4 x — 7 = ± 6.

4x = 7 ± 5 = 12 or 2.

Whence, x = 3 or -, Ans.

2

If the coefficient of x in the given equation is even, frac-
tions may be avoided, and the rule modified, as follows :

Multiply both members by the coefficient of x^, and add to
each the square of half the coefficient of x in the given equation.

2. Solve the equation 15 x^ + 28 a; = 32.
Multiplying both members by 15,

152x2+ 15(28 x)= 480.

Adding to both members the square of 14, we have

152x2 + 15(28 x) + 142 = 480 + 196 = 676.

Extracting the square root, 15 x + 14 = ± 26.

15x = - 14 ± 26 = 12 or - 40.

4 8

Whence, x = - or — , Ans.

6 3

EXAMPLES.
Solve the following equations :

3. ar'-7a; = 30. 6. 8x'-\-Ux = -3.

4. 2ic2-|-5a; = 18. 7. 10a;- + 7x = -l.

5. 3x2-2a; = 33. 8. 5a:2-2x = 72.

QUADRATIC EQUATIONS. 231

9. 4:X^-7x = -3. 14. 6a^ + 17a; = -10.

10. ex'-llx^lQ 15. 5a;=' + 15 = 28x.

11. 4a;2^ 24a; + 35 = 0. 16. 9 ar' = 32 a; - 15.

12. 4a; + 4 = 15a^. 17. 3-5a;-12a;2 = 0.

13. 4-15a;-4a;2 = 0. 18. Oar' + 15a; + 4 = 0.

MISCELLANEOUS EXAMPLES.

263. The following equations may be solved by either of
the preceding methods, preference being given to the one
best adapted to the example under consideration.

1 ^_? = §^ 3 J 1^=_1.

■ 3 2 6* ■ 8a^ 24a; 2*

2 .6x' 12a;' ■4 a; 10*

5. (3 a;+ 2) (2 a; + 3) = (a; -3) (2 a; -4).

6. 9 (a; - 1)2 - 4 (a; - 2)2 = 44.

7. 4(a;-l)(2a;-l) + 4(2a;-l)(3a;-l)

+ 4(3a;-l)(4a;-l) = 53a;2.

8 30 30 ^-^ jQ a; + 2 4-a;^7

a;a; + l * a; — 1 2a; 3

9 _J^ 2_^^ .J a; + l . 2.T + 1^17

x-Q x-5 ' ' 2a; + l 3a; + l 12*

" . 12. (2 a; + 1)2 - (3 a; - 2)^ - (a; + 1)^ = 0.

13. V6 + 10a;-3a;2^2x-3. 17 a; x" - 6 ^6

2-3a; 4-. 11 ' ' ^^^+^) ^

14.

4 a; -2 4 18. V^T2 + V3a; + 4 = 8.

15. (a;-3)3-(a;+2)3=-65. ^^ ^^j^^ 3

16. V^TH" + V3 a; + 3 = 4. 5 2+Vi2"

232 ALGEBRA.

20. ^ar^ + 8 a-^ + 1 6 X - 1 = X + 3.

21. i+_Ii?i_+ ^^! = 0.

22 1 4 _ 2

a; -6 3(x-l) 3a;

ft„ 4cc — 9 2x — 3_Q

24 ^ + ^ _L ^-^ = 6 a; + 16
■ a; -2 a; + 2 3a;

12

25. V5 + a; + V5 — x =

Vo — a;

26. Va; + 3 — Va; + 8 = — 5Vx.

27. ^.+ 1 1 ^

28. 3

1 — x^ 1 4- X 1 — a;
1 3

a: + 2 2(2 a; -3) (x + 2)(2a;-3)

I oq 3a; — 1 _ 5 —4a; _ o

7-a; 2x+l~

3Q 3a;-6 7 11 -2a;

- a; 2 2(5 - 2 a;)

«, 3-2x 2 + 3a;^l 16a; + a;^

• 2 + x 2-x 3a;2-4'

32 a; + l a; + 2^2a; + 13
a; — 1 a; — 2 a; + l

33. ^2x' + 9x + 9-{-V2x^ + 7x + 5 = V2.

34. V3a; + l-V4a; + 5 + Va;-4 = 0.

1 15a; X

35

(3a;+l)(l-5a;) 2(1 -5 a;) (7 a; +1) (3a;+l)(7a;+l)

36 V^ _ Va; + 2 _ 5^
Va; + 2 V^ 6

QUADRATIC EQUATIONS. 233

37. V5 -2x + Vl5 -3x= V26 -5x.

38 ^^ . ^±3_2(^+4}^0
a; + 3 a;-l x-2

264. Solution of Literal Quadratic Equations.

For the solution of literal affected quadratic equations,
the methods of § 262 will be found in general the most
convenient.

1. Solve the equation o^ -\- ax — bx — ab = 0.
The equation may be written

x^ + {a — h)x = ab.

Multiplying both members by 4 times the coefficient of x'^^

4x2 + 4(a - fe)x = 4a6.

Adding to both members the square of a — 6,

4 x2 + 4 (a - 6)x + (a - &)2 = 4 a& + a2 _ 2 ai + Ifi

= a2 + 2 a6 + h"^.
Extracting the square root,

2x + (a-5) = ±(a + b).

2x =-(a-6)±(a + &).

Therefore, 2x = -a + b + a + h = 2h,

or 2x = — af6 — a — 6 = — 2 a.

Whence, x = 6 or —a, Ans.

Note. If several terms contain the same power of x, the coeffi-
cient of that power should be enclosed in a parenthesis, as shown
in Ex. 1.

2. Solve the equation (m — l)oi? — 2m^x = —A.im?.
Multiplying both members by to — 1,

(m - 1)2x2 - 2 to2(to - l)x = -4 TO2(m - 1).
Adding to both members the square of to2,

(m - 1)2x2 _ 2 m2(TO - l)x + m* = TO* - 4 wi^ + 4 m2.

234 ALGEBRA.

Extracting the square root,

(m — l)x - m2 = ± (m- — 2 m).

(m — l)x = m^ + TO^ — 2 m or 7^2 _ jji2 _|. 2 TO

= 2m{m — 1) or 2 m.

2 wi

Whence, a; = 2 ni or -, Ans.

m—\

EXAMPLES.
Solve the following equations :

3. a;'-4aa; = 96--4al 6. cf? -{- ax -\- Ix + db = 0.

4. ic^ + 2 ma; = 2 ?n + 1. 1. a? — m?x — m^x = — ni\

5. x^ — (a — l)x = a. 8. acx^ + 6ca; — ada; = hdc

9. a;2-2ax-12aj = 3a2_16a-35.

10. {a-h)3?-{a-\-h)x = -2b.

11. (a - »)(a2 + &2 + ace) = a^ + hx".

12. l+^ + l=A=i. 13. _i_+-^=2.

1 — ax 1 + ax X — a x — b

14. (x + 2ay-{x-Say = 65a'.

15. (l-a2)(a; + a)-2a(l-af)=0.

16. V(a -2b)x + Sab = x + 4.b.

17. 6a^-(5a + &)a; = -a2-a6 + 2&2.

18. VaT+Ta + Vx — 2 a = V5a; + 2a.
,_ a;— a a;+a 5 aa; — 3a — 2 „

19. —^ h^ r, --2 =0.

X -]- a X — a a- — or

20. ^/x-12ab = ^^'~^''

21 a;- + 1^2(a^4-&^ 22 ^ + ^ i ^ + ^ ^ 5
a; a^ - 62 • ' x + 6 a; + a 2

23. •\/2a;- 4a +V5a; + 3a = ^^Jl^-

V2^ — 4a

QUADRATIC EQUATIONS. 235

24. x^-(a-b)x = (a- c){b - c).

/

25. V3 X + 2a — V4 x — (Ja = V2a.

\x + ?»y V^ + *'*'/

oy a; 4- Vl2a — x _ Va + 1
X — Vl2a — a; Va — 1

28. (a-6)a^ + (&-c)a; + (c-a)=0.

29. (a* -l)a^-2(a*4-l)aj = -«* + !.
30 ^ + l--<^ — ^ I c

X c a — b

31. J_ + _l_ = l + l.

a;— a x —■ b a b

32. (c + a - 2 6)x-2 +(a + 6 — 2 c)a; + 6 + c - 2a = 0.

265. Solution of Q,uadratic Equations by a Formula.

It was shown in § 261 that, if aaf + bx = c, then

^^-b±V¥+i^^
2a

This result may be used as a formula for the solution of
any quadratic equation in the form axr -{-bx = c.

1. Solve the equation 2 a;^ + 5 a; = 18.

In this case, a = 2, 6 = 5, and c = 18 ; substituting in (1),

4 4 4 2

2. Solve the equation 110 a;^ — 21 a; = — 1.

In this case, a = 110, b = — 21, and c = — 1 ; substituting in (1),

^^2l±VmZM = ^A±l = ± or 1, ^n..
220 220 10 11

Note. Particular attention must be paid to the signs of the coeffi-
cients in making the substitution.

236 ALGEBRA.

EXAMPLES.

Solve the following equations :

3.

2x' + x^6.

10.

28a;2 + 16a; = -l.

4.

if2-5a; = 36.

11.

8 x2 + 41 .-» + 5 = 0.

5.

a^ + 14 a; + 48 = 0.

12.

16a;2 + 16a.--5 = 0.

6.

5 a^- 13 a; = -6.

13.

30a;-8 = 25a;2.

7.

6 a;^ — aj = 5.

14.

12a^ + 7 = -25a;.

8.

3a^-7a; = 20.

15.

2-3a;-54a^ = 0.

9.

4a^-21a; = -27.

16.

3 + 14a;-24a^ = 0.

266. Solution of Equations by Factoring.

Let it be required to solve the equation
(a;-3)(2a; + 5)=0.

It is evident that the equation will be satisfied when cs
has such a value that one of the factors of the first member
is equal to zero ; for if any factor of a product is equal to
zero, the product is equal to zero.

Hence, the equation will be satisfied when x has such
a value that either

a; - 3 = 0, (1)

or 2a; + 5 = 0. (2)

5
Solving (1) and (2), we have a; = 3 or — -•

It will be observed that the roots are obtained by placing
the factors of the first member separately equal to zero, and
solving the residting equations.

267. 1. Solve the equation .-r — 5 a; — 24 = 0.

Factoring the first member, (x - 8) (x + 3) = 0. (§ 100)

Placing the factors separately equal to zero (§ 266), we have

X - 8 = 0, and x + 3 = 0.
Whence, x = 8 or — 3, Ans. ^

QUADRATIC EQUATIONS. 237

2. Solve the equation 2 or — a; = 0.

Factoring the first member, x(2x — 1) = 0.

Placing the factors separately equal to zero,

X = 0, and 2x — 1 = 0.

Whence, x = 0 or -, Ans.

2

J. Solve the equation x*^ 4- 4 cc^ — cc — 4 = 0.

Factoring the first member, (x + 4) (x- — 1) = 0. (§ 93)

Therefore, x + 4 = 0, and x'- — 1 = 0.

Whence, x = — 4 or ±1, Ans.

4. Solve the equation a^ — 1 = 0.

Factoring the first member, (x - l)(x2 + x+ 1) = 0. (§ 103)

Therefore, x - 1 = 0, and x^ + x + 1 = 0.

Solving the equation x — 1 = 0, we have x = 1.
Solving the equation x^ + x + 1 = 0, we have

X = -1^^^-^ (§ 265) =^J-itV33.

EXAMPLES.

Solve the following equations :

5. a;2 + 3a;-28 = 0. 10. 3 or' + 24 a;^ = 0.

6. x'-Ux + 4.5 = 0. 11. 16 a.-3 - 9 a; = 0.

7. a^ + lla; + 24=0. 12. (2aj+ 5)(9a;2 -49)= 0.

8. x2-6x-72 = 0. 13. 12ar'-7x2_io^^O

9. 5ar^-7a; = 0. 14. (ar^ - 8)(a;2 + 4)= 0.

15. (a:-3)(2.T2 + 13x + 20)=0.

16. (x - S)(x + 4)(a; - 5) - 60 = 0.

17. (ar^ - 9 a') (2 x' -\- ax - a') = 0.

238 ALGEBRA.

18. af' + l = 0. 22. 8a;3_^ 125 = 0.

19. a^-27 = 0. 23. a;^ - 64 = 0.

20. 16 a;^- 81 = 0. 24. x^-x' + x-l^O.

21. 27a^-6Aa' = 0. 25. Var^ _ V2¥^ = a; - 1

26. 5 of' -a^- 125 a; + 25 = 0.

27. 8a;3 + 20a^-18a;-45 = 0.

28. 4a;3_^ 53,2.^ 72a; + 90 = 0.

29. Va + a: + Va — a; = V^'.

30. Va + Va;- Va- Vx= V^.

Note. The above examples are illustrations of the important prin-
ciple that the degree of an equation indicates the number of its roots ;
thus, an equation of the third degree has three roots ; of the fourth
degree, four roots ; etc.

It should be observed that the roots are not necessarily unequal ;
thus, the equation x-— 2 x + 1 := 0 may be written (x — 1) (x — 1) = 0,
and therefore its two roots are 1 and 1.

PROBLEMS.

268. 1. A man sold a watch for .f 21, and lost as many
per cent as the watch cost dollars. What was the cost ?

Let X = the number of dollars the watch cost.

Then, x — the per cent of loss,

and X X -^ , or ^— = the number of dollars lost.

100 100

3.2

By the conditions, = x — 21.

100

Solving, X = 30 or 70.

Then, the cost of the watch was either $ 30 or $ 70 ; for either of
these answers satisfies the conditions of the problem.

2. A farmer bought some sheep for $ 72. If he had
bought 6 more for the same money, they would have cost
him $ 1 apiece less. How many did he buy ?

QUADRATIC EQUATIONS. 239

Let X = the number bought.

72
Then, — = the number of dollars paid for one,

72
and = the number of dollars paid for one if there

* "^ had been 6 more.

72 72

By the conditions, — = \- 1.

X X + 6

Solving, X = 18 or — 24.

Only the positive value of x is admissible, for the negative value
does not satisfy the conditions of the problem.
Therefore, the number of sheep was 18.

Note 1. In solving problems which involve quadratics, there will
usually be two values of the unknown quantity; and those values only
should be retained as answers which satisfy the conditions of the
problem.

Note 2. If, in the enunciation of the problem, the words "6
more" had been changed to "6 /eroer," and "$1 apiece less" to
"SI apiece more,''^ we should have found the answer 24.

In many cases where the solution of a problem gives a negative
result, the wording may be changed so as to form an analogous prob-
lem to which the absolute value of the negative result is an answer.

3. I bought a lot of flour for $ 126 ; and the number
of dollars per barrel was ^ the number of barrels. How
many barrels were purchased, and at what price?

4. Divide the number 18 into two parts, the sum of
whose squares shall be 170.

5. Find two numbers whose difference is 7, and whose
sum multiplied by the greater is 400.

6. Find three consecutive numbers whose sum is Aqual
to the product of the first two.

7. Divide the number 20 into two parts such that one
is the square of the other.

8. Find two numbers whose sum is 7, and the sum of
whose cubes is 133.

240 ALGEBRA.

9. Find four consecutive numbers sucli that if the firsii
two be taken as the digits of a number, that number is
equal to the product of the other two.

10. A merchant bought a quantity of flour for $ 108. If
he had bought 9 barrels more for the same money, he would
have paid $2 less per barrel. How many barrels did he
buy, and at what price ?

11. A farmer bought a number of sheep for $378.
Having lost 6, he sold the remainder for $ 10 a head more
than they cost him, and gained $ 42. How many did he
buy?

12. A merchant sold a quantity of wheat for $ 56, and
gained as many per cent as the wheat cost dollars. What
was the cost of the wheat ?

13. If the product of three consecutive numbers be
divided by each of them in turn, the sum of the three
quotients is 74. What are the numbers ?

14. A crew can row 8 miles down stream and back again
in 4f hours ; if the rate of the stream is 4 miles an hour,
find the rate of the crew in still water.

15. A certain farm is a rectangle, whose length is three
times its width. If its length should be increased by 20
rods, and its width by 8 rods, its area would be trebled. Of
how many square rods does the farm consist ?

16. A man travels 9 miles by train. He returns by a
train which runs 9 miles an hour faster than the first, and
accomplishes the entire journey in 35 minutes. Required
the rates of the trains.

17. The area of a rectangular field is 216 square rods,
and its perimeter is 60 rods. Find its length and width.

18. At what price per dozen are eggs selling when, if the
price were raised 5 cents per dozen, one would receive twelve
less for a dollar ?

QUADRATIC EQUATIONS. 241

19. A merchant sold goods for $ 18.75, and lost as many
per cent as the goods cost dollars. What was the cost ?

20. A man travelled by coach 6 miles, and returned on
foot at a rate 5 miles an hour less than that of the coach.
He was 50 minutes longer in returning than in going.
What was the rate of the coach ?

21. A square picture is surrounded by a frame. The side
of the picture exceeds by an inch the width of the frame ;
and the number of square inches in the frame exceeds by 124
the number of inches in the perimeter of the picture. Find
the area of the picture, and the width of the frame.

/ 22. The circumference of the fore- wheel of a carriage is
less by 4 feet than that of the hind-wheel. In travelling
1200 feet, the fore-wheel makes 25 revolutions more than
the hind-wheel. Find the circumference of each wheel.

23. A tank can be filled by two pipes running together
in 3f hours. The larger pipe by itself will fill it sooner
than the smaller by 4 hours. What time will each pipe
separately take to fill it ?

24. The telegraph poles along a certain railway are at
equal intervals. If there were two more in each mile, the
interval between the poles would be decreased by 20 feet.
Find the number of poles in a mile.

25. A and B gained in trade f 2100. A's money was in
the firm 15 months, and he received in principal and gain
$ 3900. B's money, which was f 5000, was in the firm 12
months. How much money did A put into the firm ?

26. If $ 2000 amounts to $ 2205, when put at compound
interest for two years, the interest being compounded annu-
ally, what is the rate per cent per annum ?

27. A man travelled 105 miles. If he had gone 4 miles
more an hour, he would have performed the journey in 9|
hours less time. How many miles an hour did he go ?

242 ALGEBRA.

28. The sum of f 120 was divided becween a certain
number of persons. If each person had received $ 7 less,
he would have received as many dollars as there were
persons. Required the number of persons.

^fSC My income is $ 5000. After deducting a percentage
for income tax, and then a percentage, less by one than that
of the income tax, from the remainder, the income is reduced
to $ 4656. Find the rate per cent of the income tax.

30. A man has two square lots of unequal size, together
containing 13,325 square feet. If the lots were contiguous,
it would require 510 feet of fence to embrace them in a
single enclosure of six sides. Find the area of each lot.

31. A merchant has a cask full of wine, containing 36
gallons. He draws a certain number of gallons, and then
fills the cask up with water. He then draws out the same
number of gallons as before, and finds that there are 25
gallons of pure wine remaining in the cask. How many
gallons did he draw each time?

32. A set out from C towards D at the rate of 5 miles an
hour. After he had gone 32 miles, B set out from D towards
C, and went every hour Jy of the entire distance ; and after
he had travelled as many hours as he went miles in an
hour, he met A. Required the distance from C to D.

33. A courier travels from P to Q in 12 hours. Another
courier starts at the same time from a place 24 miles the
other side of P, and arrives at Q at the same time as the
first courier. The second courier finds that he takes half
an hour less than the first to accomplish 12 miles. Find
the distance from P to Q.

34. A man bought a number of f 50 shares, when they
were at a certain rate per cent premium, for f 4800 ; and
afterwards, when they were at the same rate per cent dis-
count, sold them all but 30 for $ 2000. How many shares
did he buy, and how much did he give apiece ?

QUADRATIC EQUATIONS. 243

XXIII. EQUATIONS SOLVED LIKE QUAD
RATIOS.

EQUATIONS IN THE QUADRATIC FORM.

269. An equation is said to be in the quadratic form when
it is expressed in three terms, two of which contain the
unknown quantity, and the exponent of the unknown quantity
in one of these terms is twice its exponent in the other; as,

0^-6x^ = 16',

ar' + ic^ = 72 ; etc.

270. Equations in the quadratic form may be readily
solved by the rules for quadratics.

1. Solve the equation a;*' — 6 a^ = 16.

Completing the square by the rule of § 258,

x6 _ 6 a;3 + 32 = 16 + 9 = 25.
Extracting the square root, x^ — 3 =: ± 5.
Whence, ipS = 3 ± 5 = 8 or - 2.

Extracting the cube root, a; = 2 or — v'2, Ans.

Note 1. There are also four imaginary roots, which may be ob<
tained by the method of § 267.

2. Solve the equation 2x + 3 V^ = 27.

Since Vx is the same as x^, this is in the quadratic form.
Multiplying by 8, and adding 3'^ to both members (§ 261),

16x + 24 Vx + 32 = 216 + 9 = 225.

Extracting the square root, 4\/x + 3 = ± 15.

4Vx = - 3 ± 15 = 12 or - 18.

/- 9

"Whence, Vx = 3 or - -•

81
Squaring, x = 9 or — , Ans.

244 ALGEBRA.

3. Solve the equation 16 x~^ — 22 x~^ = 3.
Multiplying by 16, and adding IP to both members,

1Q"X~^ - 16 X 22 x~^ + 112 = 48 + 121 = 169.

_3

Extracting the square root, 16a; ^ — 11=±13.

16x"^ = 11 ± 13 = - 2 or 24.

_3 1 ^

Whence, a; ^ = — - or -.

8 2

i
Extracting the cube root, x~^ = — ^ or (-\ .

4
—1 1 / ^\ 3^

Raising to the fourth power, x = — - or I - ) •

4

Inverting both members, a; = 16 or ( ^ V Ans.

p
Note 2. In solving equations of the form x« = a, first extract the

root corresponding to the numerator of the fractional exponent, and

then raise to the power corresponding to the denominator. Particular

attention should be paid to the algebraic signs ; see §§ 186 and 193.

EXAMPLES.
Solve the following equations :

4. a;^-21a;2^_io8. 8. 12a;-2 + a;-i = 35.

5. 8a; + 14V^ = 15. 9. aj^ + a;'=702.

6. a^-3a;* = 88. 10. 16x-« -30.^-^=- 81.

r. 32a^ + ^ = -33. 11. 7^^-^ = 20.

ar Va;

12. (2a^-3)2 = -26a^ + 153.

13. (5 x-^ - 2)2 - 16(x-' + 1)' = - 70.

14. 9 o;"^- 22 a;"* = -8. 17. a.-^- 970;" + 1296 = 0.

in 2n

15. 3.-5 +2.^ = 16. 18. 3.t-6i = 94.
18. a;~^-34x"* = -225. a;^

QUADRATIC EQUATIONS. 246

19. 4a:-5+27iB"s" = 40. 22. 2 a;-^ + 59 a;"^ = 160.

20. 8a;~3_35a;-'2=_27. ^ , , ^ , ,

21. 27 V^ + 10 V a? = 128. ' Vx ~ Va'

24. Vs+ViK + Vs-ViK

V5+^

25. V3+V^+V4-V^'=V7 + 2Vcc.

271. An equation may sometimes be solved with refer-
ence to an expression, by regarding it as a single quantity.

1. Solve the equation (x — 5)'' — 3(x — 5)^ = 40.
Multiplying by 4, and adding 3^ to both members,

4(x - 5)3 - 12(x - 5)^ + 32 = 160 + 9 = 169.
Extracting the square root, 2(x — 5) ^ — 3= ± 13.

2(x - 5)^ = 3 ± 13 =: 16 or - 10.

3
Whence, (a; — 5) ^ = 8 or — 5.

1 3/-

Extracting the cube root, (x — 5) ^ = 2 or — v5.

Squaring, x — 5 = 4 or \/25.

Whence, x = 9 or 5 + V2E, Ans.

Certain equations of the fourth degree may be solved by
the rules for quadratics.

2. Solve the equation a;* + 12 a:^ + 34 ar^ - 12 « - 35 = 0.
The equation may be written

(x4 + 12x3 + 36x2)- 2a:2- 12x = 35.
That is, (x2 + 6 x)2 - 2 (x2 + 6 x) = 35.

Completing the square,

(x2 + 6 x)2 - 2(x2 + 6 x) + 1 = 36.
Extracting the square root,

(x2 + 6x)-l =±6.

x2 + 6x = l±6 = 7or-6.

246 ALGEBRA.

Completing the square, x^ + 6 x + 9 = 16 or 4.
Extracting the square root, x + 3 = ±4or ±2.
Whence, x = -3±4or -3i:2

= 1, — 7, — 1, or — 6, Ans.

Note 1. In solving equations like the above, we first form a per-
fect square with the x* and x^ terms, and a portion of the x^ term.
By § 258, the third term of the square is the square of the quotient
obtained by dividing the x^ term by twice the square root of the x*
term.

3. Solve the equation x^ — 6x + 5Vx- — 6 a; + 20 = 46.
Adding 20 to both members.

(x2 _ 6 X + 20) + 5 Vx^ - 6 X + 20 = 66.
Multiplying by 4, and adding 5^ to both members.

4(x2 _ 6x + 20) + 20\/x-^-6x + 20 + 52 = 264 + 25 = 289.
Extracting the square root,

2Vx2-6x + 20 + 5=±17.

2v'x2-6x + 20 = - 5 ± 17 = 12 or - 22.

Whence, Vx2-6x + 20 = 6 or - 11.

Squaring, x^ - 6 x + 20 = 36 or 121.

Completing the square, x2 _ 6 a; + 9 = 25 or 110.

Extracting the square root, x — 3=±5or ± VllO.

Whence, x = 8, - 2, or 3 ± VTlO,

Ans.

Note 2. In solving equations like the above, add such a quantity
to both members that the expression without the radical in the first
member may be the same as that within, or some multiple of it.

EXAMPLES.

Solve the following equations :

4. (x" -2xy-lS(:if-2x) = - 45.

5. x* + Sa^-10x--104:X-\- 105 = 0.

6. a;*-10x-3 + 23a^ + 10a;-24 = 0.

QUADRATIC EQUATIONS. 247

7. ^2 + 7 + Vx' + 7 = 20.

8. V3a;-2-5A/3a;-2 = -6.

9. a^-2a; + 6Var^-2x + 5 = ll.

11. A/3iB-2ar^--v/3cc-2ar^ = 2.

12. (a;3 ^ 17)1 _ 35 (^r^ ^ ij^f = _ 2I6.

13. (2 1C + 5)-^ + 31 (2 a; + 5)-'- = 32.

14. cc* + 14 x-3 + 71 x2 + 154 a; + 120 = 0.

15. 2a^-3x + 6V2ar' - 3arf2 = 14.

16. 4a;*-12ar' + 7a^ + 3a;-2 = 0.

17. (3a^ + aj-l)3-26(3ar^ + a;-l)^ = 27.

18. 4 a^ - 9 a; + 23 = 7 V4 a;2 - 9 a; + 11.
(19. (a^ + by = 2 av? + 2 a&^a; - a^a^.

20. (a;-a)3--5&(a;-a)'+6&- = 0.

21. 2 (a^ - 2 a;) + 3 Va:" - 2 a; + 6 = 15.

22. 3 a;2 - 9 a; = 4 Var* - 3 a; + 5 - 11.

23. 8(5a;-3)-^-6(5a;-3)-^ = -l.
/^4. 2 (2 af' + 10)-^ + 3 (2 a;3 _|. ^q)"^ = 2.

25. x« + 4aa;3_34ci2^_76^3^^105a* = 0.

^:

248 ALGEBRA.

XXIV. SIMULTANEOUS EQUATIONS.

INVOLVING QUADRATICS.

272. An equation containing two unknown quantities If
said to be symmetrical with respect to them when they can
be interchanged without destroying the equality.

Thus, the equation cc^ — xy + y^ = 3 is symmetrical with
respect to x and y ; for on interchanging x and y, it becomes
y^ — yx -\- a? = 3, which is equivalent to the first equation.

But the equation x — y =1 is not symmetrical with
respect to x and y ; for on interchanging x and y, it becomes
y —x =1, ora; — y = — 1, which is a different equation.

273. An equation containing two unknown quantities is
said to be homogeneous when the terms containing the
unknown quantities are of the same degree with respect
to them (§ 157).

Thus, the equation x^ — Sxy — 2y^=l is homogeneous,
for the terms containing x and y are of the second degree
with respect to x and y.

But the equation a^ — 2 ?/ = 3 is not homogeneous ; for a^
is of the second degree, and 2 y of the first degree.

274. On the use of the double signs ± and T •

If two or more equations involve double signs, it will be
understood that the equations can be read in two ways;
first, reading all the upper signs together ; second, reading
all the lower signs together.

Thus, the equations x = ±2, y = ±3, can be read either

x = -\-2, y = -\-3, or cc = — 2, 2/ = — 3.
Also, the equations x = ±2, y = ^ 3, can be read either
x = + 2, y = — 3, or x = — 2, y = + 3.

SIMULTANEOUS EQUATIONS. 249

275. Two equations of the second degree (§ 158) with
two unknown quantities will generally produce, by elimina-
tion, an equation of the fourth degree with one unknown
quantity ; the rules already given are, therefore, not suffi-
cient to solve all cases of simultaneous equations of the
second degree with two unknown quantities.

Consider, for example, the equations

x--fy=5. (1)

aj+r = 3. (2)

From (1), 2/ - 5 - cb2.

Substituting in (2), cc + 25 - 10 x-- + a;^ = 3 ;
which is an equation of the fourth degree.

In several cases, however, the solution may be effected
by means of the rules for quadratics.

276. Case I. When each equation is in the form,

ax^ + h'tf = c.

r3x2+ 42/2 = 76. (1)

1. Solve the equations [^y._^^^^^^ ^2)

Multiplying (1) by 3, 9 a;2 + 12 2/2 = 228.

Multiplying (2) by 4, 12 y"^ -4ix^= 16.

Subtracting, 53 z^ = 212.

Then, x^ = 4, and x = ± 2. (3)

Substituting from (3) in (1), 12 + 4y^ = 76.

4 j/2 = 64.
Then, y2 _ ig, and y = ± 4.

Ans. X = 2, 2/ = ± 4 ; or, x = — 2, y = ±4.

Note. In this case there are four possible sets of values of x and
y which satisfy the given equations :

1. x = 2, 2/ = 4. 3. x = -2, 2/ = 4.

2. X = 2, ?/ = — 4. 4. X = — 2, 2/ = — 4.

It would be incorrect to leave the result in the form x = ± 2,
y = ± 4 ; for, by § 274, this represents only the first and fourth of
the above sets of values.

250 ALGEBRA.

EXAMPLES.

Solve the following equations :

4:0^+ y' = ei. r 8a^- 112/2 = 8.

2ar» + 32/2 = 93. ' 1 12 ar* + 13 / = 248.

5ar-9y2 = _121. r a^+y^=5(a'-\-b^).

0.

z:^

7 2/2 -30.-^ = 105. l4rB2-2/2=5a(3a-4&).

277. Case II. When one equation is of the second degree,
and the other of the first.

Equations of this kind may always be solved by finding
the value of one of the unknown quantities in terms of the
other from the simple equation, and substituting this value
in the other equation.

1. Solve the equations |2a^- a^y = 6^/. (1)

^ 1 x + 2y = l. (2)

From (2), 2y = l -x, or y= 1^^. (3)

Substituting in (1), 2«2 _ xH-^^^, = G^^^^l^V

Clearing of fractions, 4x2 — 7x + x2 = 42 — Gjc

Or, 6a;2_a; = 42.

Solving this equation, a; = 3 or — --.

5

7 + li
7 — 3 5 49

Substituting in (3), y = — - — or — - — = 2 or — •

Z it lU

14 49

X = 3, « = 2 ; or, X = ,y — — , Ans.

Mf , , 5 ^ 10

Note. Certain examples where one equation is of the third degree,
and the other of the first, may be solved by the method of Case II.

EXAMPLES.
Solve the following equations :

2.

■5ar=-32/2 = -7.
2x+y = 7.

xy = — 54.

SIMULTANEOUS EQUATIONS.

251

4.
5.
6.

7.

9.

x-y = l.
a^ + y' = 113.

x' + xy — y^ = — 19.
x-y = -l. --.-^

a^-f = - 117.
x-y = -3.

3^ + f = 217.
x + y=7.

x-y = l.
xy = a^ + a.

4
3'

3 4

10 f^ + 2/ = 2a.

I a;2 + y2 ^ 2 (a'' + fc').

jj rar^ + 27/ = 98.
|a; + 3y = 2.

^ + ^ = 1.
X y

12.

13.

(li

15.

« — 3?/ = 5.

1 1^5_
X y 2

2x + 2y=^5.

7 .^2 + 10 .r?/ = - 8.
5x + 4?/ = — 8.

2 + ^ = 12.
y a; 3

3a:-22/ = -12.

278. Case III. When the equations are symmetrical with
respect to x and y (§ 272), and one equation is of the second
degree, and the other of the second or first.

Equations of this kind may always be solved by combin-
ing them in such a way as to obtain the values oi x + y
and x — y.

1. Solve the equations

ix + y = 2. (1)
\xy = -15. (2)

Squaring (1),

x^ + 2 xy + y2 = 4.

Multiplying (2) by 4,

4 a-y = - 60.

Subtracting,

x^-2zy + 2/2 = 64.

Extracting the square root,

x-y = ±%. (3)

Adding (1) and (3),

2« = 2±8 = lOor -6.

Whence,

a; = 5 or — 3.

Subtracting (3) from (1),

2?/ = 2 t8 = -6 or 10.

Whence,

y = — 3 or 5.

sc = 5,

y = - 3 ; or, z = - 3, y = 5, Am.

252 ALGEBRA.

Note 1. In subtracting ± 8 from 2, we have 2 ^ 8, in accordance
with the notation explained in § 274. In operating with double signs,
± is changed to T , and T to ± , whenever + would be changed to — .

Note 2. The above equations may also be solved by the method
of Case II. ; but the symmetrical method is shorter and neater.

Certain examples in which one equation is of the third
degree, and the other of the first or second, may be solved
by the method of Case III.

0^-^ = 56. (1)

2. Solve the equations , „

^ 'x-2 + x?/ + ?/2=28. (2)

Dividing (1) by (2), x - y = 2. (3)

Squaring (3), x^ -2xy + y^ = 4. (4)

Subtracting (4) from (2), 3xy = 24, or xy = 8. (5)

Adding (2) and (5), x'^ + 2xy + y'^ = 36.

Whence, x + y = ±6. (6)

Adding (3) and (6), 2« = ±6 + 2 = 8or-4.

"Whence, x = 4 or — 2.

Subtracting (3) from (6), 2 y = ± 6 - 2 = 4 or - 8.

"Whence, y = 2 or — 4.

X = 4, y = 2 ; or, x = — 2, y = — i. Ans.

Note 3. The above equations are not symmetrical according to
the definition of § 272 ; but the method of Case III. may often be used
in cases where the given equations are symmetrical except with respect
to the sigtis of the terms.

x' + y^ = 50. (1)

3. Solve the equations ,

^ ^ xy = -7. (2)

Multiplying (2) by 2, 2xy = - 14. (3)

Adding (1) and (3), x'^ + 2xy + y'^ = 36.

Whence, x + y = ±6. (4)

Subtracting (3) from (1), x'^-2xy + y^ = 64.
Whence, x — y = ± 8. (5)

Adding (4) and (5), 2x = 6 ± 8, or - 6 ± 8.

Whence, x = 7, — 1, 1, or — 7.

Subtracting (6) from (4), 2 y = 6 T 8, or - 6 ^ 8.

Whence, y = — 1, 7, — 7, or 1.

x = ±7,y=^l; or, X = ± 1, y = T 7, Ans.

SIMULTANEOUS EQUATIONS. 253

EXAMPLES.

Solve the following equations :

4. \^^ = ''\, 12. f-' + 2'' = 260.

[ ic + 2/ = 14. [x — y = —14:.

Ja;== + / = 101. .„ lxy = -SO.

^- \x + y = -9. "• \x-y = 24.

U^_f = 87. lx^ + f = 504.

^ \x^ + xy + f-^37. ■ \x'-xy + f = S4.

7.

xy = 45. .. f. i af — xy + y^ = 63.

X —y = ~ 4. ' \ X — y = — 3.

(xy = 12. (x' + f = 305.

°- iar^ + / = 40. [a; -2/ = 21.

ra.^_^^133. I ^.-2 + 2/2 = 218.

"^ \x-y = 7. '■ \xy = -91.

10. (^ + 3/^ = -217. ^3_ I ^ + ^ = -335^

[ cc + 2/ = — < . t ^ — a;?/ + 2/- = 67.

^^- U + y = -2. ^^- \x-y = -31.

279. Case IV. TF^e^i each equation is of the second degree,
and homogeneous (§ 273).

Note 1. Certain equations which are of the second degree and
homogeneous may be solved by the method of Case I. or Case III.
(See Ex. 1, § 276, and Ex. .3, § 278.)

The method of Case IV. should be used only when the example
cannot be solved by the methods of Cases I. or III.

i. Solve the equations J „ ~ '

^ \x'-\- y^ = 29.

Putting y = vx in the given equations, we have

a;2 - 2 t;x2 = 5 ; or, x^ = — ^ — ; (1)

1 — 2 V

and x" + v2x2 = 29 ; or, x2 = ^^

1 +U2

* Divide the first equation by the second.

254

Equating the values of a;^,

Or,
Or,
Solving this equation,

ALGEBRA.

5 29

1 - 2 V 1 + ^2
5 + 5v2 = 29-58v.
5i;2 + 58v = 24.

V =- or — 12.
5

Substituting these values in (1), x^ =

1-^
5

S or 5

rT24 = 25or25

Whence,

x = ±5 or ±

V5

Substituting the values of v and x in the equation y = vx,

If v = - andx = ±5, ?/=?(±5) = ±2.
6 5

If u = — 12 and x = ±

V5

, 2/ = -12(

, V5\ ^12V5

2.

^ns. x = ±5, 2/ = ±2; or, x=±-\/6, y = ^--V5.

Note 2. In finding y from the equation y = vx, care must be
taken to multiply each pair of values of x by the corresponding value
of u.

EXAMPLES
Solve the following equations :
- 2 ar' - a;y = 28.
. a^ + 2 2/2 = 18.

' x' -{- xy = — 6.
xy — y^ = — 35.

x^ -\- xy -\- y^ = 63.

cb2 _ 2/2 ^ _ 27.

x' + 3y' = 28.

aj2 + a;2/ + 2?/2=16.

a^ - 2 a;?/ = 84.
2xy — y^ = — 64.

4.

6.

8.

9.
10.
11.

3x^ + xy-3y^ = 33.
2x''-y' = 23.

x^ -\- 5xy — y^ = — 7.
lar'-f-3a;?/-2/ = -4.

(x'-xy-12y- = S.
[x' + xy-10y' = 20.

(5a^-4:xy=z33.

'. 27 a^-32a;2/-4 2/2=55.

■3x^-{-xy + y^ = 47.
, 4 a;' — 3 ajy — 2/2 = — 39.

SIMULTANEOUS EQUATIONS. 255

MISCELLANEOUS AND REVIEW EXAMPLES.

280. No general rules can be given for the solution of
examples which do not come under the cases just considered.
Various artifices are employed, familiarity with which can
only be gained by experience.

x^-f= 19. (1)

1. Solve the equations , „ „ ^
'x'y-xf- = 6. (2)

Multiplying (2) by 3, 3 x^y -3xy^ = 18. (3)
Subtracting (3) from (1), x^ -Sx^y + 3x2/2 -y^ = l.

Extracting the cube root, x — y = 1. (4)

Dividing (2) by (4), xy = 6. (5)

Solving equations (4) and (6) by the method of Case III., we find
X = S, y — 2 ; or, x = — 2, y = — 3, Ans.

( cc^ -\- y^ = 9 xy.

2. Solve the eq'.iations \

Putting X = « + V and y = u — v, we have

(u + vy + (u-vy = 9(u+v)(u-v), (1)
and (u+ «) + (?< -v) = 6. (2)

Reducing (1), 2 «» + 6 uV^ = 9(«a _ v^). (3)

Reducing (2), 2 w = 6, or « = 3.

Substituting the value of u in (3), 64 + 18«2 = 9(9 - v"^).
Whence, v^ = 1, or « = ± 1.

Therefore, a; = M + v = 3±l=4or2,

and ?/ = n — v = 3Tl = 2or4.

X = 4, y = 2 ; or, X = 2, y = 4, Ans.

Note. The artifice of substituting u + v and m — r for a; and y is
applicable in any case where the given equations are symmetrical with
respect to x and y (§ 272). See also Ex. 4, p. 256.

3. Solve the equations

x' + f-{-2x-{-2y = 23. (1)

.xy = 6. (2)

Multiplying (2) by 2, 2 xy = 12. (3)

Adding (1) and (3), x^ + 2xy + y"^ + 2x + 2y= 35.

256 ALGEBRA.

Or, (x + 2/)2 + 2(a; + 2/)=35.

Completing the square, (x+t/)2+2(a;+y) + l = 36.

Whence, (.x + y)+ 1 =±6,

or

X + y = 5 or — 7. (4)

Squaring (4), x^ + 2xy + y^ = 25 or 49.

Multiplying (2) by 4, 4 xy = 24.

Subtracting, x^ — 2xy + y'^= 1 or 25.

Whence, x — y = ±lor±d. (5)

Adding (4) and (5), 2x = 5 ± 1, or — 7 ± 5.

Whence, x = 3, 2, - 1, or - 6.

Subtracting (5) from (4), 2?/ = 5 q= 1, or — 7 T 5.

Whence, y = 2, 3, - 6, or - 1.
r=3, y=2 ; x=2, y=3 ; x= — 1, y=—6; or, x= —6, y= — 1, Ans.

( X* -I. y* := 97

4. Solve the equations J " "^ -^ *

[ x-\-y = — l.

Putting x = u + V "nd y = u — v, we have

(m + vy + (u- vy = 97, (1)

and (u + v) + (u-v) = -l. (2)

Reducing (1), 2 m* + 12 uH^ + 2 v* = 97. (3)

Reducing (2), 2 m = — 1, or u — — -.

Substituting in (3) , 1 + 3 ^2 + 2 «* = 97.
8

25 SI

Solving this equation, v2 _ _ or — ' —

4 4

Whence, „ = ± § or ± ^~ ^^.

2 2

Then,

x = « + .=-i±^or-l±^^HlI=2, -3, orHli^^^M,
2 2 2 2 2

and

, = „_, = _l:F5or-lT^^^II = -3,2, orZLlTV^I.
^ 2 2 2 2 2

x=2, r/=-3; x=-3, y = 2; or, ^^-1^^-31^ ^ ^ - 1 .p v^ZIsT^

SIMULTANEOUS EQUATIONS.

257

EXAMPLES.
Solve the following equations :

8.

10.

^ + 2/- = l.

^ 25

a^ + / 4- a; - y = 26.
xy = 12.

2ar-3a^ = -4.
4:xy — 5y- = 8.

4: a:^ — 5 xy = 19.
xy-{-y^=&.

\x y 2

i- = _J_.
xy 18*

0^2 + 2 2/2 = 47 _,_ 2 a;.

a^-2/ = -7.

11. |ar' + a?y + 2/' = 97.
[ a; — ?/ = 19.

12.

13.

14.

15.

:i? + f = 756.
x^-xy + y^ = 63.

iry + 28 a;y - 480 = 0.
2x + y = n.

i+i

a^
1_1

^ 2/

r

_5^
16*

( y^ -f- v' = 17.
\x + y=l.

16.

17.

18.

19.

20.

21.

22.

23.

' a? — y^ = ^.
xy = -2.

ra:2 + 4/-f-3aj = 22
[2xy + 3y-{-9 = 0.

\ 3x^ — 5xy -\-2 f- ztT.- 8
[4x-53/ = 10.

xy

= a?-l.

24.

25.

26.

x + y =:2a.

^ + 4 = 91.

1 + 1 = 7.

» 2/

a^+y 1 ^ — y — 12.

X — ?/ x + y 3
a^ + ^2 ^ 45.

ra^ + 2/' = 2a^ + 6a&2.
[xy{x + y)^2o?-2ah''

2x'-3xy=\5a-10a?.
3x + 2y=12a-13,

a;" + a^/ + / = 9i
a;2 + xy 4- y2 = 13.

ar^ + 2/2 = 13 (a^ + l).
aj4-2/ = 5a — 1.

2 0^^+3 0;^ — 42/2=_2C
5a^-72/2=_8.

* Divide the first equation by the second.

258

ALGEBRA.

27.

28.

29.

30.

31.

a

32.

33.

34.

35.

36.

x" ~ xy + y^ = 3a' - Sab + 3b\

(x'-\-y* =
\x-y =

* = 97.
5.

' 9x^ ~ xy — y = 51.
- 5 .Tt/ + / -f- 3 a; = 81.

X — y = 19.
■^x — -\/y = l.

y X 2

x-\-y = l.

37.

38.

39.

40.

41.

f = 3a^ + 3a + l.^

3?

x-y = l.

f 2(?y — a; = — 14.
( a;y + 0^ = 148.

x^ — xy =■ 27 y.
xy — y^ = 3x.

42.

43.

44.

(x + y ^ 2x

y

x — y x-{-2y
[x~3y = — 2.

y{x — a)=2 ab.
x{y — 6)= 2ab.

15
4

= -T' 45.

46.

a;5 _ 2^ = 31.

a; - 2/ = 1.

x^ = X -{- y.
y'^ = 3y-x.

r Va^ + 7 = 6 - y.

1 V»*T227 = 22 - a^.

53 a^- 128 a;?/ +64 2/2 =5.
26a^-62a;i/ + 32/=5,

xy — (x — y) = 1.

a;?/+ (a;--?/)=-5.
a^yC^J — y)=-84.

x^ — xy -\-y^ = 12.
a:3 + 2/3 + 3a;?/ = -48.

■ x^ + ajy + y2 = 7.
a; + y = 5 + a;y.

' 2 ar^ + 2 2/^ = 5 a^.
ar* + / = 33.

X* — 2 x?/ + 3.TZ = —16.
2a;-3?/ = 7.
I 3 a? + 52! = -14.

PROBLEMS.
Note. In the following problems, as in those of § 268, only those
answers are to be retained which satisfy th'' given conditions.

281. 1. The sum of the squares of two numbers is 52,
and their difference is one-fifth of their sum. Find the
numbers.

2. The diiference of the squares of two numbers is 16,
and their product is 15. Find the numbers.

SIMULTANEOUS EQUATIONS. 259

3. If the leugth of a rectangular field were increased by
2 rods, and its width diminished by 5 rods, its area would
be 80 square rods ; and if its length were diminished by 4
rods, and its width increased by 3 rods, its area would be
168 square rods. Pind its length and width.

4. The difference of the cubes of two numbers is 218, and
the sum of their squares is equal to 109 minus their prod-
uct. Find the numbers.

5. If the product of two numbers be multiplied by their
sum, the result is 70 ; and the sum of the cubes of the num-
bers is 133. Find the numbers.

6. A farmer bought 4 cows and 8 sheep for $ 600. He
bought 5 more cows for f 490 than sheep for $80. Find
the price of each.

I 7. Find a number of two figures such that, if its digits be
inverted, the difference of the number thus formed and the
original number is 9, and their product 736.

8. The sum of two numbers exceeds the product of their
square roots by 7 ; and if the product of the numbers be
added to the sum of their squares, the result is 133. Find
the numbers.

9. The sum of the terms of a fraction is 13. If the
numerator be decreased by 2, and the denominator increased
by 2, the product of the resulting fraction and the original
fraction is -^. Find the fraction.

10. A rectangular mirror is surrounded by a frame 3^
inches wide. The area of the mirror is 384 square inches,
and of the frame 329 square inches. Find the length and
width of the mirror.

11. A crew row up stream 18 miles in 4 hours more time
than it takes them to return. If they row at two-thirds their
usual rate, their rate up stream would be one mile an hour.
Find their rate in still water, and the rate of the stream.

V

260

ALGEBRA.

12. A rectangular field contains 2\ acres. If its length
were (decreased by 10 rods, and its width by 2 rods, its aiea
would be less by an acre. Find its length and width.

13. A distributes f 180 equally among a certain number
of persons. B distributes the same sum between a number
of people less by 40, and gives to each $ 6 more than A does.
How many persons are there, and how much does A give
to each ?

14. A, B, and C together can do a piece of work in one
hour. B does twice as much work as A in a given time;
and B alone requires one hour more than C alone to per-
form the work. In what time could each alone do the work ?

15. If the length of a rectangular field were increased by
one-eighth of itself, and its width decreased by one-sixth of
itself, its area would be decreased by 60 square rods, and its
perimeter by 2 rods. Find its length and width.

16. If the product of two numbers be added to their
difference, the result is 26; and the sum of their squares
exceeds their difference by 50. Find the numbers.

(Represent the numbers hy x + y and x — y.)

17. A sets out to walk to a town 21 miles off, and one
hour afterwards B starts to follow him. When B has over-
taken A, he turns back, and reaches the starting-point at
the same instant that A reaches his destination. B walked
at the rate of 4 miles an hour. Find A's rate, and the dis-
tance from the starting-point to where B overtook A.

18. A tank can be filled by three pipes, A, B, and C>
when opened together, in 2j?_. hours. If A filled at the
same rate as B, it would take 3 hours for A, B, and C to
fill the tank ; and the sum of the times required by A ami
C alone to fill the tank is double the time required by B
alone. In what time can each pipe alone fill the tank?

19. The sum of two numbers is 4, and the sum of theii
fifth powers is 244. Find the numbers.

THEORY OF QUADKATIC EQUATIONS. 261

XXV. THEORY OF QUADRATIC EQUA-
TIONS.

282. Sum and Product of the Roots.

Let Ti and Vi denote the roots of the equation x^ -\- px = q.

-D COCK -i)+Vp^ + 4g — p-Vp^ + 4^
By § 265, rj = —^ ^ ~' ^^^ **2 = —^ j ^•

— 2»
Adding these values, ri + r2 = — = —p.

Multiplying them together, we have

p^ - (/>^ + 4 q) — 4 g

TiTi = ^-^ ^ (§ 80) = —^ = -q.

Hence, if a quadratic equation is in the form aP -\- px = q,
the sum of the roots is equal to the coefficient of x with its sign
changed, and the product of the roots is equal to the second
member with its sign changed.

1. Find the sum and product of the roots of the equation
20?*- 7a; -15-0.

Transposing — 15, and dividing by 2, the equation becomes

x2 - li = — .

2 2

7 15
Hence, the sum of the roots is -, and their product .

EXAMPLES.

Find by inspection the sum and product of the roots of :

2. x' + 7x-^6 = 0. 6. 12a;2-4a: + 3 = 0.

3. ar'- a; + 12 = 0. 7. 9a:-21ar + 7 = 0.

4. a;2 + 3a;-l = 0. 8. 4 -a; -6x^ = 0.

5. 3ic2_a;_6 = 0. 9. 14ar' + 8ax + 21a2 = 0.

262 ALGEBRA.

283. Formation of Equations.

By aid of the principles of § 282, a quadratic equation
may be formed which shall have any required roots.
For, let Ti and r^ denote the roots of the equation

x'+px-q=0. (1)

Then by § 282, j? = — ^i — rj, and — g = r^r^

Substituting these values in (1), we have

a^ —TyX — r^ + r-^Ti = 0.

That is, {x - r^ {x - r^) = 0. (§ 93)

Hence, any quadratic equation can be written in the form

(x-r,)ix-r,) = 0, (2)

where r^ and r^ are its roots.

Therefore, to form a quadratic equation which shall have
any required roots.

Subtract each of the roots from x, and place the product of
the resulting expressions equal to zero.

1. Form the quadratic equation whose roots shall be 4

and

4

By the rule,

(a;-4)(x + ^) = 0.

Multiplying by 4,

(a;-4)(4x + 7) =0.

"Whence,

4x2 -9x- 28 = 0, Ans.

EXAMPLES.
Form the quadratic equations whose roots shall be ;

2. 6, 9.

-^'-1

5 6

8.

-f'»-

3. 2, - 3.

5.-.-f.

» 5 3
7- - f 4

9.

5 8
"4'~9

THEORY OF QUADRATIC EQUATIONS. 263

10. 2a-\-b,a-3b. 12. 3 + 7 V2, 3 - 7 V2.

11. a + 3m, a-3m. 13. h-Va-\-Vb),h-Va-Vby

FACTORING.
284. Factoring of Quadratic Expressions.
A quadratic expression is an expression of the form

ax^ + bx + c.

The principles of § 283 serve to resolve such an expres-
sion into two factors, each of the first degree in x.

We have, aa^ + bx + c = afa^ -h — ^-X (1)

\ a aj

Now let ri and rj denote the roots of the equation

a a,
By § 283, (2), the equation can be written in the form
(x — ri) (x — r^) = 0.

Hence, the expression a^ -\ \-~ can be written

a a

(x - ri) (x - ra).

Substituting in (1), we have

Oic* -\- bx + c = a{x — ri) (x — r^.

But Ti and r^ are the roots of the equation o^ + — -f- £ = 0,

a a
or aa? + 6a; + c = 0 ; which, we observe, is obtained by placing
the given expression equal to zero.

We then have the following rule :

To factor a quadratic expression, place it equal to zero, and
solve the equation thus formed.

Then the required factors are the coefficient of x^ in the given
eaapression, x minus the first root, and x minus the second root.

264 ALGEBRA.

EXAMPLES.

285. 1. Factor 6 cc^ + 7 x - 3.

Solving the equation 6 a:^ + 7 « — 3 = 0, we have by § 265,

^ ^ - 7 ± V49 + 72 ^ - 7 ± 11 ^ 1 ^^ _3
12 12 3 2'

Then by the rule, 6x^+ 1 x - 3 = 6(x --\ (x + -\

= (3x- l)(2a; + 3), Ans.
2. Factor 4. + 13x-12x^
Solving the equation 4 + 13 a; — 12 a;^ = 0, we have by § 266,

- 13 ± \/l69 + 192 - 13 ± 19 1 4

X = — — ^ = — = or —

-24 -24 4 3

Whence, 4 + 13x- 12x2 = - 12[x + lVx --^

= 4(a= + l)x(-3)(x-|) ■

= (l + 4x)(4-3x), Ans.
Factor tlie following :

3. a;2_i3a; + 42. 14. Qx" -23mx + 21m\

^. x' + Wx-^U. ' 15. Uc^ + 25x + 6.

5. a^-9a;-36. 16. 18 cc^ - 15 a; + 2.

6. 3ar= + 7a;-6. 17. 5-19a;-4a^.

7. 5ar^+18a; + 16. 18. 18 ar' + 31 a; + 6.

8. Gx'-llx-^-S. 19. 45 + 7 a; - 12 ar'.

9. 15a^-14aj-8. 20. 42 + 23 a; - 10 ar^.

10. 20-7a;-:a^. 21. 2A x" - 26 x -\- 5.

11. 35-lla;-6ar'. 22. 8a^ + 38a; + 35.

12. 12 + 28a;-5a;2. 23. 21 ar' - 10 a^ - 24 /.

13. 3x'-nax-28a\ 24. 7 x" + 37 abx - 30 a'b^.

r

THEORY OF QUADRATIC EQUATIONS. 265

25. Factor 2x^-3xy~2y^~7x-\-4:y + 6.
Placing the expression equal to zero, we have

2x^-Sxy -2y^-7x + iy + G = 0,
or 2x2-(32/ + 7)x = 2y2_4y_6.

Solving this by the formula of § 265,

_3y + 7±A/(3y + 7)-^+ 16y^-32y-48
X- -

^Sy + 7±V2by^ + 10y + i^3y + 7±(5y+l)
4 4

8y + 8 -2y + 6 „ „ -y + S
- ^ or ^-^— = 2j^ + 2or ^

4 4 ' ' 2

Therefore,

2 x2 - 3 a:?/ - 2 2/2 _ 7 X + 4 2/ + 6 = 2 [x - (2 ?/ + 2)] Tx - ~ ^ J" ^1

= (x-2?/-2)(2x + 2/-3), ^?is.
Factor the following :

26. x- + xy-12y'-\-7x + 7y-\- 12.

27. x'^-xy-2y^-{-x-5y-2.

28. rc^- 42/- 4-3 a; + 10?/ -4.

29. 2^2 + 7a;?/ -4/ + a; + 137/ -3.

30. 3 a^ - 5 a6 - 2 62 _ 7 a + 2.

31. 6 - 15 ?/- 5a; + 9?/2 + 9a;?/ -4a^.

32. 6a.'2-9a;?/ + a;z-15?/2_13?/z-2 22.

286. If the coefficient of x^ is a perfect square, it is con-
venient to factor the expression by the artifice of completing
the square (§ 260) in connection with § 99.

1. Factor 9 a^ - 9 a; - 4.

By § 260, the expression 9 x^ — 9 x will become a perfect square by

9 3

adding to it the square of ■ — -, or — Then,

2\/9 ^

(r-:

9a;2_9a;_4 = 9a;2_9x + (;^]---4 = ^3x-^y-?^

266 ALGEBRA.

Factoring as in § 99, we have

9x^-9x-4 = (3.-| + |)(3x-?-|)

= (3a;+ l)(3x-4), Ans.

If the x^ term is negative, the entire expression should be
enclosed in a parenthesis preceded by a — sign.

2. Factor 3 - 12 « - 4 a;^.

3 -12a; -4x2 = -(4x2+ 12x-3)

= _ (4 x2 + 12 X + 32 - 9 - 3)

= -[(2x + 3)2-12]

= (2x + 3 + \/l2)x(- l)(2x + 3-\/l2)

= (2V3 + 3 + 2x)(2V3-3-2x), Ans.

EXAMPLES.
Factor the following :

3. x'-5x + 4:. 9. 36a^ + 24a^-5.

4. 4a;2 + 16a; + 15. 10. 4.x' + 5x-6.

5. 9a;2-18x' + 8. 11. 25 ar= + 30 a; + 6.

6. 16 a:^ + 16 a; - 21. 12. 4 +12 a; -9x1

7. ar' + 2.'B-ll. 13. 49^2 + 56a; + 12.

8. 4x2 + 4a;-l. 14. 5 + 38 a; - 16 x^.

287. Certain trinomials of the form ax* + bx^ + c, where
a and c are perfect squares, may be resolved into two fac-
tors by the artifice of completing the square.

1. Factor 9 a;" - 28 .t^ ^ 4

By § 96, the expression will become a perfect square if its middle
term is — 12 x2.

Thus, 9x*-28x2 + 4=(9x*- 12x2 + 4)- IGx^
= (3x2-2)2-(4x)2

= (3x2-2 + 4x)(3x2-2-4x) (§99)

= (3x2 + 4x-2)(3x2-4x-2), Ans.

\

THEORY OF QUADRATIC EQUATIONS. 267

2. Factor a' + a'^b^ + b\

a* + a%^ + 6* = (a* + 2 a'-b'^ + ¥) - a'^lP-
= {a?- + y'-y - a-^62
= (a2 + &2 + a6)(a2 + ft'^ - a6)
= (a2 + a& + 62) (a2 _ at + 6-^), Ans.

3. Factor a;^ + 1-

a;4 + 1 =(a:4 + 2x2+ l)-2a;2

= (x2 + l)2_(x\/2)2

= (x2 + a; v/2 + 1) (ar.2 - x \/2 + 1), Av.s.

EXAMPLES,
Factor the following :

4. a;* + 2a^ + 9. 12. a;" + 16.

5. a;* -19x2 + 25. 13. a!*-5a^ + l.

6. 4 a* + 7 a2&2 _^ 16 6^ 14. 9 a* - 55 aV _f. 25 3^4^

7. 9a;''-28a;V + 4?/^ 15. 16 a* + 47 aW + 36 m*.

8. 16 m^ - m-V + ri*. 16. 25 x* - 21 a^ + 4.

9. 4«^-53a2 + 49. 17. 25 m* + 36 mV + 16 a;*.

10. 9a;* + 5a;2 + 9. 18. 16 a;* - 60 3.^ + 49?/*.

11. 4m*-13m2 + 4. 19. 36 a* - 68 a^ft^ ^ 25 6^

288. Certain equations of the fourth degree may be
solved by factoring the first member by the method of
§ 287, and then proceeding as in § 267.

1. Solve the equation a;* + 1 = 0.

By Ex. 3, § 287, the equation may be written

(x2 + a; \/2 + 1) (x2 - x V2"+ 1) = 0.
Then, as in § 267, x2 + x \/2 + 1 = 0, and x2 - x a/2 + 1 = 0.
Solving the equation x2 + x V2 + 1 = 0, we have by § 265,

^ _ -V2±y2^^ _ - \/2 ± \A^

268 ALGEBRA.

Solving the equation x- — x V2 + 1 = 0, we have

V2 ± •\/2 -4 \/2 ± V^

X = = = ~ •

2 2

EXAMPLES.
Solve the following :

2. ic* - 2 a'2 + 25 = 0. 5. cc^ + a^ + l = 0.

3. a;*-18ar + 9 = 0. 6. a;* -9 a;- + 9 = 0.

4. 4ic''-5a^ + l = 0. 7. a:^ + 81 = 0.

DISCUSSION OF THE GENERAL EQUATION.

289. By § 265, the roots of the equation oc? -\- px = q are

' 1 — 2 ' 2 — 2

We will now discuss these values for all possible real
values of p and q.

I. Suppose q positive.

Since p"^ is essentially positive (§ 186), the expression
under the radical sign is positive, and greater than pi

Therefore, the radical is numerically greater than p.

Hence, rj is positive, and r^ is negative.

If p is positive, r, is numerically greater than rj ; that is,
the negative root is numerically the greater.

If j9 is zero, the roots are numerically equal.

If p is negative, ?-i is numerically greater than r^', that
is, the positive root is numerically the greater.

II. Suppose q = 0.

The expression under the radical sign is now equal to pi
Therefore, the radical is numerically equal to p.
If p is positive, Vi is zero, and r2 is negative.
If p is negative, Vi is positive, and rg is zero.

THEORY OF QUADRATIC EQUATIONS. 269

III. Suppose q negative, and 4g mimerically <2r.

The expression under the radical sign is now positive, and
less than pi

Therefore, the radical is numerically less than p.
If |) is positive, both roots are negative.
If p is negative, both roots are positive.

IV. Suppose q negative, and 4 q numerically equal to p-.
The expression under the radical sign now equals zero.
Hence, Vi is equal to rg.

If p is positive, both roots are negative.
If p is negative, both roots are positive.

V. Suppose q negative, and 4 q numerically > p^.

The expression under the radical sign is now negative.
Hence, both roots are imaginary (§ 248).
The roots are both rational or both irrational, according
as p^ + 4 g is or is not a perfect square.

EXAMPLES.

290. 1. Determine by inspection the nature of the roots
of the equation 2 or — 5 x — 18 = 0.

The equation may be written x^ = 9 ; here p = and q = 9.

Since q is positive and p negative, the roots are one positive and
the other negative ; and the positive root is numerically the greater.

In this case, p^ + iq = — +S6= ; a perfect square.

4 4

Hence, the roots are both rational.

Determine by inspection the nature of the roots of the
following :

2. 6a;2 + 7a;-5 = 0. 7. 16a;--9 = 0.

3. 10a^ + 17iK + 3 = 0. 8. 9ar-l = 12a;.

4. 4a^-a; = 0. 9. 25 a:- + 30 ic + 9 = 0.

5. 4.a^- 20 x + 25 = 0. 10. 7x- + 3x = 0.

6. a;^ - 21 x + 200 = 0. 11. 41 x = 20 ic^ _^ 20.

270 ALGEBRA.

XXVI. ZERO AND INFINITY.

VARIABLES AND LIMITS.

291. A variable quantity, or simply a variable, is a quan-
tity which may assume, under the conditions imposed upon
it, an indefinitely great number of different values.

A constant is a quantity which remains unchanged
throughout the same discussion.

292. A limit of a variable is a constant quantity, the dif-
ference between which and the variable may be made less
than any assigned quantity, however small, but cannot be
made equal to zero.

In other words, a limit of a variable is a fixed quantity
to which the variable approaches indefinitely near, but
never actually reaches.

Suppose, for example, that a point moves from A towards
B under the condition that it shall move, during succes-
sive equal intervals of time,

first from A to C, half-way f f f f T

between A and B; then to

D, half-way between C and B ; then to E, half-way between

D and B; and so on indefinitely.

In this case, the distance between the moving point and
B can be made less than any assigned quantity, however
small, but cannot be made equal to zero.

Hence, the distance from A to the moving point is a vari-
able which approaches the constant value AB as a limit.

Again, the distance from the moving point to -B is a
variable which approaches the limit 0.

293. A problem is said to be indeterminate when the
rnunber of solutions is indefinitely great. (Compare § 159.)

ZERO AND INFINITY. 271

294. Interpretation of -•

Consider the series of fractions

a a a a
3' ^3' X)3' X)03'"**'
where each denominator after the first is one-tenth of tlie
preceding denominator.

It is evident that, by sufficiently continuing the series,
the denominator may be made less than any assigned quan-
tity, however small, and the value of the fraction greater
than any assigned quantity, however great.

In other words,

If the numerator of a fraction remaiyis constant, while the
denominator approaches the limit 0, the value of the fraction
increases without limit.

It is customary to express this principle as follows :

a

- = CO.
(J

Note. The symbol oo is called Infinity.

295. Interpretation of — •

■^ GO

Consider the series of fractions

3' 30' 300' 3000' '
where each denominator after the first is ten times the pre-
ceding denominator.

It is evident that, by sufficiently continuing the series,
the denominator may be made greater than any assigned
quantity, however great, and the value of the fraction less
than any assigned quantity, however small.

In other words,

If the numerator of a fraction remains constant, while the
denominator increases without limit, the value of the fraction
approaches the limit 0.

272 ALGEBRA.

It is customary to express this principle as follows.
«=0.

00

296. It must be clearly understood that no literal meaning
can be attached to such results as

- = 00, or — = 0;
0 ' 00 '

for there can be no such thing as division unless the divisor
is a Jinite quantity.

If such forms occur in mathematical investigations, they
must be interpreted as indicated in §§ 294 and 295. (Com-
pare note to § 395.3

THE PROBLEM OF THE COURIERS.

297. The discussion of the following problem will serve

tu further illustrate the form -, besides furnishing an inter-

0
pretation of the form -•

The Problem of the Couriers. Two couriers, A and B,
are travelling along the same road in the same direction,
BE', at the rates of m and n miles an hour, respectively.
If at any time, say 12 o'clock, A is at P, and B is a miles
beyond him at Q, after how many hours, and how many
miles beyond P, are they together ?

B p q ar

I LI L i

Let A and B meet x hours after 12 o'clock, and y miles
beyond P.

They will then meet y — a miles beyond Q.

Since A travels mx miles, and B nx miles, in x hours, we
have

j y = mx.
\y — a = nx.

ZERO AND INFINITY. 273

Solving these equations, we obtain

a J am

X = , and y =

m — n m — n

We will now discuss these results under different hypoth-
eses.

1. m ^ n.

In this case, the values of x and y are positive.

Hence, the couriers will meet at some time after 12
o'clock, and at some point to the right of P.

This corresponds with the hypothesis made ; for if m is
greater than m, A is travelling faster than B ; and it is evi-
dent that he will eventually overtake him at some point
beyond their positions at 12 o'clock.

2. m < n.

In this case, the values of x and y are negative.

Hence, the couriers met at some time before 12 o'clock,
and at some point to the left of P. (Compare § 10.)

This corresponds with the hypothesis made ; for if m is
less than n, A is travelling more slowly than B; and it is
evident that they must have been together before 12 o'clock,
and before they could have advanced as far as P.

3. m = n, or m — w = 0.

In this case, the values of x and y take the forms - and

0

am ,■ 1

— , respectively.

If m — w approaches the limit 0, x and y increase with-
out limit (§ 294) ; hence, if m = n, no finite values can be
assigned to x and y, and the problem is impossible.

Thus, a result in the form - indicates that the problem is
impossible.

This interpretation corresponds with the hj^Dothesis made ;
for if m = n, the couriers are a miles apart at 12 o'clock,
and are travelling at the same rate ; and it is evident that
they never could have been, and never will be together.

274 ALGEBRA.

4. a = 0, and m > n or m < n.

in this case, a; = 0 and y — 0.

Hence, the couriers are together at 12 o'clock, at P.

This corresponds with the hypotliesis made ; for if a = 0,
and m and n are unequal, the couriers are together at 12
o'clock, and are travelling at unequal rates ; and it is evi-
dent that they never could have been together before 12
o'clock, and never will be together afterwards.

5. a = 0, and vi = n.
in this case, the values of x and y take the form -.

If a = 0, and m = n, the couriers are together at 12 o'clock,
and are travelling at the same rate.

Hence, they always have been, and always will be, together.

In this case, the number of solutions is indefinitely great;
for any value of x whatever, together with the correspond-
ing value of y, will satisfy the given conditions.

Thus, a 7'esult in the form - indicates that the problem is
indeterminate (§ 293).

THE THEOREM OF LIMITS.

298. If tioo variables are alioays equal, and each approaches
a limit, the limits are equal.

AM C B A' Mf Bf

I I I 1 I \ \

Let AM and A!M' be two variables which are always
equal, and approach the limits AB and AB\ respectively.

If possible, suppose AB > AB\ and lay off AG — AB\

Then the variable AM may assume values between AC
and AB, while the variable AM is restricted to values less
than AQ\ which is contrary to the hypothesis that the
variables should always be equal.

Hence AB cannot be > AB\ and in like manner it may
be proved that AB cannot be < AB' ; therefore AB = AB.

INDETERMINATE EQUATIONS- 275

XXVII. INDETERMINATE EQUATIONS.

£t was shown in § 159 that a single equation which
contains two or more unknown quantities is satisfied by an
indefinitely great number of sets of values of these quanti-
ties. If, however, the unknown quantities are required to
satisfy other conditions, the number of solutions may be
finite.

We shall consider in the present chapter the solution of
indeterminate equations of the first degree, containing two
or more unknown quantities, in which the unknown quanti-
ties are restricted to positive integral values.

299. Solution of Indeterminate Equations in Positive
Integers.

1. Solve the equation 1 x -{-by = 118 in positive integers.

Dividing by 5, the smaller of the two coefficients, we have

5 0

Or, 'Ix-Z = 23 - X - ?/.

5

Since, by the conditions of the problem, x and y must be positive

2 X .3

integers, it follows that must be an integer.

5

Let this integer be represented by p.

Then, 2x-Z ^^^^ or 2 x - 3 = Sp. (i)

5

Dividing (1) by 2, x-l-i = 2^+--
Or, X - 1 - 2j) =^-=ti.

Since x andp are integers, x-l-2j9 is also an integer ; and there-

•e ^ "*" must be an integer.
2

Let this integer be represented by q.

276 ALGEBRA.

Then, ^-^ = q, or p = 2q-l.

Substituting in (1), 2a;-3 = 10g-5.

Wlience, 2 x = 10 g - 2, and a; = 6 g - 1. (2)

Substituting tliis value in the given equation,

35g-7 + 5?/ = 118.
Whence, 5 y = 125 - 35 g, and ?/ = 25 - 7 g. (3)

Equations (2) and (3) form what is called the general solution in
integers of the given equation.

Now if g is zero, or any negative integer, x will be negative ; and if
g is any positive integer greater than 3, y will be negative.

Hence, the only positive integral values of x and y which satisfy the
given equation are those arising from the values 1, 2, 3 of g.

If g = 1, a; = 4, and y — 1%; if g = 2, a; = 9, and y=^l\; if g = 3,
X = 14, and y — i.

2. In how many ways can the sum of $ 15 be paid with
dollars, half-dollars, and dimes, the number of dimes being
equal to the number of dollars and half-dollars together ?

Let X = the number of dollars,

y = the number of half-dollars,
and z = the number of dimes.

Then, lOx + 5^ + ^ = 150, (1)

and z = x + y. (2)

Subtracting (2) from (1),

10 X -(- 5 2/ = 150 - a; - y, or 11 x -f- 6 ?/ = 150. (3)

Dividing by 6, x + ^+ y = 2o.
G
5 X
Then, — - must be an integer ; or, x must be a multiple of 6.
6

Let X = 6p, where p is an integer.

Substituting in (3), 66p + 6 y = 150, or ?/ = 25 - 11 j9. ^

Substituting in (2), z = 6p+25- lip = 25 - 5p.

The only positive integral solutions are when p = 1 or 2; if. p = I,
X = 6, y = 14, and z = 20 ; if ;> = 2, x = 12, y = 3, and z = 15.

Then the number of ways is two ; either 6 dollars, 14 half-dollars,
and 20 dimes ; or 12 dollars, 3 half-dollars, and 15 dimes.

INDETERMINATE EQUATIONS. 277

EXAMPLES.
Solve the following in positive integers :

3. 2x + 2,y = 21. 9. 43aj + 10y = 719.

4. 7a; + 4i/ = 80. 10. 8.x + 19y = 700.

5. 7x + 38?/ = 211. (2x + ^v-Bz = -

11. ' ^

6. 31 a; + 9 ?/ = 1222. ''"• [ 5a; - t/ + 4^ = 21

7. 24 a; + 7?/ = 422. r 3a;_2y - 2 = - 57.

12. ' ^

8. 8 a; + 67 2/ = 158. •"^- [ 6a;-f lit/ + 2^ = 348.

/^olve the following in least positive integers :

13. 4a;-3y = 5. 16. 21a;- 8^/ = - 25.

14. 5a;-7y = ll. 17. 13a; - 30?/ = 61.

15. 19a;-42/ = 128. 18. 17 a; - 58 2/ = - 79.

19. In how many different ways can the sum of $ 2.10 be
paid with twenty-five and twenty-cent pieces ?

y 20. In how maiay different ways can the sum of $ 3.90 be
paid with fifty and twenty-cent pieces ?

21. Find two fractions whose denominators are 9 and 5,
respectively, and whose sum shall be equal to -^J^-.

22. In how many different ways can the sum of $ 5.10 be
paid with half-dollars, quarter-dollars, and dimes, so that
the whole number of coins used shall be 20 ?

23. A farmer purchased a certain number of .pigs, sheep,
and calves for f 160. The pigs cost f 3 each, the sheep $ 4
each, and the calves $ 7 each ; and the number of calves
Avas equal to the number of pigs and sheep together. How
many of each did he buy ?

24. In how many different ways can the sum of $ 5.45
be paid with quarter-dollars, twenty -cent pieces, and dimes,
so that twice the number of quarters plus 5 times the num-
ber of twenty-cent pieces shall exceed the number of dimes
by 36?

^

278 ALGEBRA.

XXVIII. RATIO AND PROPORTION.

300. The Ratio of one number to another is the quotient
obtained by dividing the first number by tlie second.

Thus, the ratio of a to & is - ; and it is also expressed a : b.

b

301. A Proportion is a statement that two ratios are
equal.

The statement that the ratio of a to 6 is equal to the
ratio of c to d, may be written in either of the forms

%. ^ tt c

a : 0 = c : a, or - = —
' b d

302. The first and fourth terms of a proportion are called
the extremes, and the second and third terms the means.

The first and third terms are called the antecedents, and
the second and fourth terms the consequents.

Thus, in the proportion a : b = c : d, a and d are the ex-
tremes, b and c the means, a and c the antecedents, and
6 and d the consequents.

303. If the means of a proportion are equal, either mean
is called a Mean Proportional between the first and last
terms, and the last term is called a Third Proportional to
the first and second terms.

Thus, in the proportion a:b = b:c, b is a mean propor-
tional between a and c, and c is a third proportional to
a and b.

304. A Fourth Proportional to three quantities is the
fourth term of a proportion whose first three terms are the
three quantities taken in their order.

RATIO AND PROTORTION. 279

Thus, in the proportion a : b = c : d, d is a. fourth propor-
tional to a, b, and c.

305. A Continued Proportion is a series of equal ratios,
in which each consequent is the same as the following ante-
cedent; as,

a : b = b : c = c : d = d : e.

PROPERTIES OF PROPORTIONS.

306. In any j^roportion, the product of the extremes is equal
to the product of the means.

Let the proportion be a:b = c:d.

Then by §301, . 1 = ^-

b d

Clearing of fractions, ad = be.

307. A mean proportional betiveen two quantities is equal
to the square root of their product.

Let the proportion be a:b = b : c.

Then, 6^ ^ ^^ (^^ ^qq^

Whence, b = -Vac.

308. From the equation ad = be, we obtain

be -, , ad
a = — , and b = —
d c

That is, in any proportion, either extreme is equal to the
product of the means divided by the other extreme ; and
either mean is equal to the product of the extremes divided
by the other mean.

309. (Converse of § 306.) If the product of tioo quantities
is equal to the product of tico others, one pa?V maj/ be made
the extremes, and the other pair the tneans, of a proportion.

280 ALGEBRA.

Let ad = be.

T-w- -J- I, 1 7 ad be a c

Dividing by ott, — = ^ or - = —

^ "^ ' bd bd b d

Whence by § 301, a:b = c:d.

In like manner, we may prove that

a: c = b: d,

c : d = a : b, etc.

310. In any proportion, the terms are in proportion by
Alternation; that is, the first term is to the third as the second
term is to the fourth.

Let the proportion be a: b = c: d.

Then, ad = be. (§ 306)

Whence, a:c = b:d. (§ 309)

311. In any proportion, the terms are in proportion by
Inversion; that is, the second term is to the first as the fourth
term is to the third.

Let the proportion be a : b = c : d.

Then, ad = be. (§ 306)

Whence, b:a = d:c. (§ 309)

312. In any proportion, the terms are in proportion by
Composition ; that is, the sum of the first two terms is to the

first term as the sum of the last two terms is to the third term.
Let the proportion be a:b = c:d.
Then, ad = be.

Adding each member of the equation to ac,

ac + ad = ac + be.
Or, a{c + d) — c(a -\- b).

Whence, a -t b : a ^c + dTc.' (§ 309)

In like manner, we may prove that

a + b : b — c + d : d.

RATIO AND PROPORTION. 281

313. In any proportion, the terms are in proportion by
Division; that is, the difference of the first tivo terms is to
the first term as the difference of the last two terms is to the
third term.

Let the proportion be a : b = c : d.

Then, ad = be.

Subtracting each member of the equation from ac,

ac — ad = ac — be.
Or, a(c — d) = c(a — b).

Whence, a — b : a = c — d : c.

Similarly, a — b : b = c ~ d : d.

314. In any proportion, the terms are in proportion by
Composition and Division ; that is, the sum of the first two
terms is to their difference as the sum of the last two terms
is to their difference.

Let the proportion be a:b — c:d.

Then by § 312, ^^+^ = ^±^. (1)

a

c

a — b

c — d

a

c

a + b

c-\-d

And by §313, ^^^^^^^-u,^ ^2)

Dividing (1) by (2),

a — b c — d

Whence, a -\-b : a — b = c + d: c — d.

315. In a series of equal ratios, any antecedent is to its con-
sequent as the sum of all the antecedents is to the sum of all
the consequents.

Let a:b = c: d = e:f.

Then by § 306, ad = be,

and cif= be.

Also, ab = ba.

Adding, a(b + d +/) = 6 (a + c + e).

Whence, a:6 = a + c + e:6 + d-f-/. (§309)

a _
b~

c

ma_

mb

nc
nd

282 ALGEBRA.

In like manner, the theorem may be proved for any num-
ber of equal ratios.

316. In any proportion, if the first two terms he multiplied
by any quantity, as also the last two, the resulting quantities
will be in proportion.

Let the proportion he a:b — c: d.
Then,

Therefore,

Whence, ma : mb = nc : nd.

In like manner, we may prove that

a b c d
VI ' m n' n

Note. Either to or n may be unity ; that is, either couplet may
be multiplied or divided without multiplying or dividing the other.

317. In any proportion, if the first and third terms be mul-
tiplied by any quantity, as also the second and fourth terms,
the resulting quantities ivill be in proportion.

Let the proportion be a : b = c : d.
Then,

Therefore,

Whence, ma : nb — mc : nd.

In like manner, we may prove that

ah c d
m' n m' n

Note. Either m or n may be unity.

a _
h~

c ^
' d

ma
nb

mc
nd

RATIO AND PROPORTION. 283

318. In any member of proportions, the products of the cor-
responding terms are in proportion.

Let the proportions be a : 6 = c : d,
and e :/= g : h.

Then, ^ = ^,and^ = 2.

h d f h

Multiplying these equals, we have

a e _c g ae eg

h^f~d^V "'' Vf^dJi
Whence, ae : hf—cg: dh.

In like manner, the theorem may be proved for any num-
ber of proportions.

319. In any proportion, like powers or like roots of the
terms are in projoortion.

Let the proportion he a:b = c: d.

Then, ^ = ^.

0 d ,

Therefore, ^ = 2!. ^^

&" d"

Whence, a" : 6" = c" : d".

In like manner, we may prove that

Va : "v^ = Vc : -y/d.

320. If three quantities are in continued proportion, the
first is to the third as the square of the first is to the square
of the second.

Let a : & = 6 : c.

Then, ^ = ^.

' h c

Therefore, ^ x ^ = ^ x ^, or ^ = ^'.

b c b b c Ir

Whence, a : c = a" : b'.

284 ALGEBRA.

321. If four quantities are in continued pro^iortion, the
first is to the fourth as the cube of the first is to the cube of
the second.

Let a:b = b:c = c:d.

bed

Then,

r^, r a b c a a a a a^

Therefore, t X - x -, = 7 X t X tj or - = -.

b c d b b b d b^

Whence, a:d — a^:W.

PROBLEMS.

\ 322. 1. Solve the equation

2a; + 3:2aj-3 = 26+a:26-a.

By § 314, 4x:6 = 46 :2 a.

Dividing the first and third terms by 4, and the second and fourth

terms by 2 (§ 317), we liave

X : 3 = 6 : a.

Whence by § 308, x = — , Ans.

2. If x:y={x-\- zf : {y + z)-, prove that 2; is a mean pro-
portional between x and y.

From the given proportion, y(x + zY = x(y + z)^. (§ 306)

Or, x^y + 2 xyz + yz"^ = xy'^ + 2 xyz + xz"^.

Or, x'^y — xif = xz"^ — yz'^.

Dividing by x — y, xy = z-.

Therefore, 2 is a mean proportional between x and y (§ 307).

3. If - = -, prove that

b d

a'^ — b-:a- — 3ab — c^ — d^:cr — S cd.

Let - = - = X ; whence, a = bx.
b d

--1

a-i - 62 _ h^x^ b"' _ x'^-l _ d^ ^ c"--cP

' a' -Sab ~ bH^^SbH ~ x"^ - 3 x ~ c^ _ 3c ~ c^ - 3 cd*

d^ d
Whence, o^ - 6^ : a^ - 3 a6 = c^ - d^ : c^ - 3 cd.

RATIO AND PROPORTION. 285

4. Find a fourth proportional to 35, 20, and 14.

5. Find a mean proportional between 18 and 50.

6. Find a third proportional to ^- and -j^.

7. Find the second term of a proportion whose first,
third, and fourth terms are 5-^, 4|, and 1|.

8. Find a third proportional to a^ — 9 and a — 3.

9. Find a mean proportional between 5| and ISy^^.

10. Find a mean proportional between

and — ■ •

*\ a; + 4 x + 2,

Sol^e the following-^tjuations : ~--n^

11. 5a;— 3a:5a;''+3a = 7a V5:13a-5.

12. 2x^l:^x-l = lx-{-l7t~x-3.

13. .f- - 16 : a^ - 25 = a;2 - 2 ic - 24 : rcf - 3 oj - 10.

14. 1 — vT^3.Jr+-\/l— x= V^— V6 — a: V&+V6 — a.

( ax'^by :bx + ay = a^ ~ b^ : 2ab. ^

^ [xy = a?b^. ">. , ■ i /\ ~^^

16. Find two numbers in the ratio' 16 to 9 such that, if
each be diminished by 8, they shall be in the ratio 12 : 5.

■■ 17. Divide 36 into two parts such that the greater dimin-
ished by 4 shall be to the less increased by 3 as 3 is to 2.

18. Find two numbers such that, if 4 be added to each,
they will be in the ratio 5 to 3 ; and if 11 be subtracted
from each, they will be in the ratio 10 to 3.

19. There are two numbers in the ratio 3 to 4, such that
their sum is to the sum of their squares as 7 is to 50. What
are the numbers ?

20. lilx-^z:^x-3z = 4.y-lz:^y-8z, prove that
2 is a mean proportional between x and y.

286 ALGEBRA.

21. If ma -\- nb : pa + qb = mb -{- nc : pb -\- qc, prove that
6 is a mean proportional between a and c.

22. If 2a-6:4a-hS^'^2^-d:^ + 3d, prove that
a:b — c:d.

23. If 8 cows and 5 oxen cost four-fifths as much as 9
cows and 7 oxen, what is the ratio of the price of a cow to
that of an ox ?

24. Given {a~ + ab)x + (6^ - ab)y^ (a' + b')x - {a' - b^j ;
find the ratio of x to y.

25. Find a number such that if it be added to each term
of the ratio 5 : 3, tlie result is | of what it would have been
if the same number had been subtracted from each term.

/

If - = -, prove that
b d

26. 2a + 36:2a-36 = 2c + 3d:2c-3d

27. a^-{-2ab:3ab-4:b'=c^ + 2cd:3cd-4. d\

28. a^ - a'b + ab- :a'-W = c'- c'd + cd' : c' - d\

29. The population of a town increased 2.G per cent from
1870 to 1880. The number of males decreased 3.8 per cent
during the same period, and the number of females increased
10.6 per cent. Find the ratio of males to females in 1870.

30. Each of two vessels contains a mixture of wine and
water ; in one the Avine is to the water as 1 to 3, and in the
other the wine is to the water as 3 to 5. A mixture from
the two vessels is composed of wine and water in the ratio
9 to 19. Find the ratio of the amounts taken from each
vessel.

31. The second of three numbers is a mean proportional
between the other two. The third number exceeds the sum
of the other two by 15, and the sum of the first and third
exceeds twice the second by 12. Find the numbers.

\

VARIATION. 287

XXIX. VARIATION.

^'323. One quantity is said to vary directly as another
when the ratio of any two values of the first is equal to
the ratio of the corresponding values of the second.

Note. It is customary to omit the word "directly," and say
simply that one quantity varies as another.

324. Let us suppose, for example, that a workman
receives a fixed sum per day.

The amount which he receives for m days will be to
the amount which he receives for n days as m is to n.

That is, the ratio of any two amounts received is equal to
the ratio of the corresponding numbers of days worked.

Hence, the amount which the workman receives varies as
the number of days during which he works.

*^32S. One quantity is said to vary inversely as another
when the first varies directly as the reciprocal of the second.
Thus, the time in which a railway train will traverse a
fixed route varies inversely as the speed ; that is, if the
speed be doubled, the train will traverse its route in one-
half ih.e time.

\/ 326. One quantity is said to vary as two others jointly
when it varies directly as their product.

Thus, the wages of a workman varies jointly as the
amount which he receives per day, and the number of
days during which he works.

327. One quantity is said to vary directly as a second
and inversely as a third, when it varies jointly as the
second and the reciprocal of the third.

Thus, in physics, the attraction of a body varies directly
as the quantity of matter, and Inversely as the square of
the distance.

288 ALGEBRA.

328. The symbol cc is read " varies as " ; thus, a oc 6 is
read "a varies as &."

329. If xccy, then x is equal to y multiplied by a constant
quantity.

Let x' and y' denote a fixed pair of corresponding values
of X and y, and x and y any other pair.
Then by the definition of § 323,

X y x'

-, = ^,, or x = -y.
x' y' y'

.r'
Denoting the constant ratio — by m, we have

y'

x — my.

330. It follows from §§ 325, 326, 327, and 329 that:

1. If x varies inversely as ?/, x = —

y

2. If X varies jointly as y and z, x = myz.

my

3. If X varies directly as y and inversely as z, x =

331. Problems in variation are readily solved by convert-
ing the variation into an equation by aid of §§ 329 or 330.

PROBLEMS.

332. 1. li X varies inversely as y, and is equal to 9 when
y = 8, what is the value of x when y = 1S?

If X varies inversely as y, we have a; = — (§ 330).

y

Putting X = 9 and y = 8, we obtain 9 = — , or m = 72.

8

Tlien, X = — ; and if ?/ = 18, x =— ^ = 4, Aiis.
y 18

2. Given that the area of a triangle varies jointly as its
base and altitude, what will be the base of a triangle whose
altitude is 12, equivalent to the sum of two triangles whose
bases are 10 and 6, and altitudes 3 and 9, respectively ?

Let B, H, and A denote the base, altitude, and area, respectively,
of any triangle, and B' the base of the required trifingle.

VARIATION. 289

Since A varies jointly as B and H, we have A = mBH (§ 330).
Then the area of the first triangle is m x 10 x 3, or 30 m, and the
area of the second is wi x 6 x 9, or 54 m.

Whence, the area of the required triangle is 30 m + 54 »», or 84 m.
But the area of the required triangle is also m x B' x 12.
Therefore, 12 mB' - 84 m, and B' = 7, Ans.

3. If ?/ cc X, and is equal to 40 when x = 5, what is its
value when a; = 9 ?

4. It ycc z^, and is equal to 48 when z = 4, what is the
expression for y in terms of z^ ?

5. If X varies inversely as y, and is equal to |- when
y = ^, what is the value of y when a; = f ?

6. If z varies jointly as x and y, and is equal to -| when
y = 4 and x = |, find the value of z when x = ^ and y = f .

7. If X varies directly as y and inversely as z, and is
equal to -j^ when y = 27 and 2 = 64, what is the value of
X when ?/ = 9 and z = 32?

8. If 5 a; + 8 oc 6 2/ — 1, and x = 6 when y = — 3, what
is the value of cc when y = 7 ?

9. If x*ccy^, and a; = 4 when 2/ = 4, what is the value
of y wlien x = ^?

10. The distance fallen by a body from a position of rest
varies as the square of the time during which it falls. If it
falls 257|- feet in 4 seconds, how far will it fall in 6 seconds ?

11. Two quantities vary directly and inversely as x,
respectively. If their sum equals — \^ when x = l, and
— -| when x = — 2, what are the quantities ?

12. The area of a circle varies as the square of its diame-
ter. If the area of a circle whose diameter is 4 is ^, what
will be the diameter of a circle whose area is ^^ ?

13. If the volume of a pyramid varies jointly as its base
and altitude, find the base of a pyramid whose altitude is
11, equivalent to the sum of two pyramids, whose bases are
13 and 14, and altitudes 0 and 7, respectively.

290 ALGEBRA.

14. Given that y is equal to the sum of two quantities
which vary directly as x^ and inversely as x, respectively.
If y — — \; when X = 1, and y — ^ when x — — 2, what is
the value of y when a; = — -i- ?

15. Three spheres of lead whose radii are 6, 8, and 10

inches, respectively, are melted and formed into a single
sphere. Find its radius, having given that the volume of
a sphere varies as the cube of its radius.

16. The volume of a cone of revolution varies jointly
as its altitude and the square of the radius of its base.
If the volume of a cone whose altitude is 3 and radius of
base 5 is -§^^, what will be the radius of the base of a
cone whose volume is ^^ and altitude 5 ?

17. If 7 men in 4 weeks can earn $238, how many men
will earn $ 121\ in 3 weeks ; it being given that the amount
earned varies jointly as the number of men, and the number
of weeks during which they work ?

18. If the A^olume of a cylinder of revolution varies
jointly as its altitude and the square of its radius, what
will be the radius of a cylinder whose altitude is 3, equiva-
lent to the sum of two cylinders whose altitudes are 5 and
7, and radii 6 and 3, respectively ?

19. If the illumination from a source of light varies in-
versel}^ as the square of the distance, how much farther
from a candle must a book, which is now 15 inches off, be
removed, so as to receive just one-third as much light ?

20. Given that y is equal to the sum of three quantities,
the first of which is constant, and the second and third vary
as X and a?, respectively. If y = — 19 when x = 2, y = 4
when X = 1, and y =2 when x = — l, what is the expres-
sion for y in terms of a;?

(Represent the constant by Z, and the other two quantities by niz
and nx^.)

PROGRESSIONS. 291

XXX. PROGRESSIONS.

ARITHMETIC PROGRESSION.

333. An Arithmetic Progression is a series of terms each
of which is derived from the preceding by adding a con-
stant quantity called the common difference.

Thus, 1, 3, 5, 7, 9, 11, ••• is an arithmetic progression in
which the common difference is 2.

Again, 12, 9, 6, 3, 0, — 3, ••• is an arithmetic progression
in which the common difference is — 3.

334. Given the first term, a, the common difference, d, and
the number of terms, n, to find the last term, I.

The progression is a, a-\-d, a-\-2d, a-{-Sd, •••.
It will be observed that th'e coefficient of d in any term
is 1 less than the number of the term.

Then in the nth or last term the coefficient of d is n — 1.
That is, l = a + (n - 1) d. (I.)

335. Givsn the first term, a, the last term, I, and the mvin-
her of terms, n, to find the sum of the terms, S.

S = a +{a + d)-]-(a + 2 d)-\ \-{l — d)-{-l.

Writing, the terms in reverse order,

S = I -\-{l - d) + {l - 2 d)-\- ••• -\-(a-\-d)+a.
Adding these equations term by term,
2 S = (a + l) + (a + l)-\-(a + 1)+ ... +(a-\-T)-t(a + 1).
Therefore, 2S = n(a + I), and >S' = '^ (a + l). (II.)

336. Substituting in (II.) the value of I from (I.), we
have >Sf = ^[2a+(n-l)ri].

292 ALGEBRA.

EXAMPLES.

337. 1. Find tlie last term and the sum of tlie terms of
the progression 8, 5, 2, ••• to 27 terms.

In this case, a = 8, d — 5 — 8 = — 3, and n = 27.
Substituting in (I.), Z = 8 +(27 - 1)(- 3)= 8 - 78 =- 70.
Substituting in (II.), S = —(8 - 70) = 27 x (- 31) = -837.

Note. The common difference may be found by subtracting the
first term fi'om the second, or any term from tlie next following terra.

Find the last term and the sum of the terms of :

2. 3, 9, 15, ••• to 12 terms.

3. —7, —12, —17, ••• to 15 terms.

4. -69, -62, -55, ••• to 16 terms.

5. -, --, --, ••' to 17 terms.

6. ^, ^, II, ... to 13 terms.
4' 12' 12'

7. -h i % ■■' to 22 terms.

o ^ o

8. , — '-, , ••• to 55 terms.

4 6 12

9. --, --, --, ••• to 19 terms.

5 2 5

10. 2a-5&, 6a-2&, 10a + &, ••. to 9 terms.

,, x — yyoy — x ^^^,

11. ^ ^, ^ -^ — , ••• to 10 terms.

^1 Li Li

338. If any three of the five elements of an arithmetic
progression are given, the other two may be found by sub-
stituting the given values in the fundamental formulae (I.)
and (II.), and solving the resulting equations.

PROGRESSIONS. 293

5 5

1. Given a = — -, ?i = 20, S = ; find d and I.

o o

Substituting the given values in (II.). we have

-^=lol-^ +l], or -'^ = -^+1; whence, Z=^- A = ?.
3^3^ 6 3' 362

Substituting the values of Z, a, and n in (I.), vre have

- = -- + 19d; wlience, l9d = - + -=-, and (Z = 1.
2 3' 23 6 6

2. Given d = -3, I = -39, >S' = - 264 ; find a and n.
Substituting in (I.), - 39 = a + (u - 1)(- 3), or a = 3n - 42. (1)
Substituting the values of ^S", a, and I in (II.), v?e have

-264=-(3tt-42-39), or -528=3 «2_81 n, or ?j2_27»=-176.

■n7T,^„^^ 27 ± V729 - 704 27 ± 5 ,« „

Whence, n= — — = — — — =: 16 or 11.

2 2

Substituting in (1), a = 48 - 42 or 33 - 42 = 6 or - 9.
Therefore, a = Q and ?i = 16 ; or, a =— 9 and n = 11, Ans.

Note 1. The interpretation of the two answers is as follows:
If a = 6 and n = 16, the progression is

6, 3, 0, - 3, - 6, - 9, - 12, - 15, - 18, - 21, - 24, - 27, - 30,
- 33, - 36, - 39.

If a = — 9 and n = 11, the progression is
- 9, - 12, - 15, - 18, - 21, - 24, - 27, - 30, - 33, - 36, - 39.
In each of these the sum is — 264.

3. Given a = -, ^ = — ^7, '^ = ~ o » fi^d I and m.

Substituting in (I.), Z = ^ + (n - 1) ( _ 1) = ^. (1)

Substituting the values of S, a, and I in (II.), we have

2 2U 12 J V 12 J

™^ 9 ± V81 + 144 9 ± 15 i„^^ „

Whence, n = — ^ — = —=^ — - = 12 or — 3.

2 2

36.

294 ALGEBRA.

The value n =— 3 is inapplicable, for the number of terms in a

progression must be a positive integer.

5 lo 7

Substituting the value n = 12 in (1), I =

12 12

Therefore, I — — - and n = 12, Ans.
12

Note 2. A negative or fractional value of n is inapplicable, and
must be rejected, together with all other values dependent upon it.

EXAMPLES.

4. Given d = 5, 1 = 11, 7i = 15 ; find a and S.

5. Given d = - 4, 7i = 20, aS = - 620 ; find a and Z.

6. Given a = — 9, ?i = 23, Z = 57 ; find d and iS'.

7. Given a = — 5, n = 19, aS* = — 950 ; find d and Z.

8. Given a = -, 1 = —, S = — -^ ; find cZ and n.

4 4 2

3

9. Given 1 = — -, w = 19, aS = 0 ; find a and d.

o

10. Given d = —-, S = —, « = -; find Z and n.

JLii o o

15 1

11. Given a = ^, Z = — ttt' ^~~o9' find 'i- and >S.

12. Given c? = -, « = 17, /S = 17 ; find a and L

5

13. Given Z = 6, d = -, /S = 24 ; find a and n.

6

14. Given Z = -5i, w = 21, /S = - 381 ; find a and d

5 23

15. Given a = — -, Z = ~, ^S = — 91 ; find d and n.

Li Li

16. Given a = -, n = 15, aS = ; find d and I.

4 8

17. Given a = — -, d = , S = -^—; find ?i and Z.

2 4 4

PROGRESSIONS. 295

18. Given I = —-, cZ = — -— , S = — —-; find a and n.

o 15 3

19. Given a = 5, d = , S = — S0; find 7i and I.

o

From (I.) and (II.), general farmulce for the solution of
examples like the above may be readily derived.

20. Given a, d, and S ; derive the formula for 7i.

By § 336, 2S = nl2a + (n - 1) d], or dn^ -\- (2 a - d)n =: 2 S.
This is a quadratic in n ; and may be solved by the method of § 261.
Multiplying by 4d, and adding {2 a — d)- to both members,
4 dhi^ + 4 d(2 a - d)n + (2 a - d)2 = 8 cZ/S' + (2 a - d)2.
Extracting the square root,

2dn-\-2a-d = ± y/^dS-\-(2a- d)^

Whence, n = d -2a ±^idS -,(2a - d)- ^„,.
2d

21. Given a, I, and n ; derive the formula for d.

22. Given a, n, and iS ; derive the formulas for d and l.

23. Given d, n, and S ; derive the formulae for a and I.

24. Given a, d, and I ; derive the formulae for n and yiS.

25. Given cZ, Z, and n ; derive tjie formulae for a and >S.

26. Given I, n, and /iS ; derive the formulae for a and d.

27. Given a, d, and /S^ ; derive the formula for I.

28. Given a, I, and S ; derive the formulae for d and 7i.

29. Given d, I, and ^S; derive the formulae for a and n.

339. To twseri any number of arithmetic means between
two given terins.

1. Insert 5 arithmetic means between 3 and — 5.

"We are to find an arithmetic progression of 7 terms, whose first
term is 3, and last term — 5.

296 ALGEBRA.

Putting a — S, I = — 5, and n = 7, in (I.), § 334, we have

— 5 = 3 + 6 d ; wlience, 6 d = — 8, and d =

3

Hence, the required progression is

Q 5 1 1 7 11 . .

o, -, -, —1, — , , —5, Ans.

3 3 3 3

EXAMPLES.

2. Insert 6 arithmetic means between 3 and 8.

10 5

3. Insert 4 arithmetic means between — and

3 2

4

3
3
2
5
4

4. Insert 5 arithmetic means between and 1.

3

3 9

5. Insert 7 arithmetic means between and —

2 2

6. Insert 8 arithmetic means between and — 5.

7. Insert 9 arithmetic means between - and — 11.

340. Let X denote the arithmetic mean between a and b.
Then, by the nature of the progression,

X — a = b — X, or 2x = a -{-b.

Whence, a;=^^+-^.

2

That is, the arithmetic mean betiveen two quantities is equal
to one-half their sum.

EXAMPLES.

Find the arithmetic mean between :

1. A and -A. 3. 2^^ and ^~^^-
12 20 2a + l 2a-l

2. (x + Tf and (x - If. 4. -"^^ and - ^!±1'.
^ ^ ^ ^ a-b a^-W

PROGRESSIONS. 297

PROBLEMS.

341. 1. The sixth term of an arithmetic progression is

5 16

-, and the fifteenth term is — Find the first term.

6 3

By § 334, the sixth term is a + 5 d, and the fifteenth term a + 14 d.

[«+ 5(? = |- (1)

Then by the conditions, -!

!a + 14d = |. (2)

Q \

Subtracting (1) from (2), 9d = - ; whence, d = —

Jj it

Substituting in (1), a + - = - ; whence, a = — , Ans.

2 6 3

2. Find four numbers in arithmetic progression such that
the product of the first and fourth shall be 45, and the
product of the second and third 77.

Let the numbers be x — 3 y, x — y, x + y, and x + Sy.

x2 - 9 2/2 = 45.
2/2 = 77.
Solving these equations, x=9, 2/=±2 ; or, x=— 9, y =±2 (§ 276).
Then the numbers are 3, 7, 11, 15 ; or, — 3, — 7, — 11, — 15.

f X2 —

Then by the conditions, < „

Note. In problems like the above, it is convenient to represent
the unknown quantities by symmetrical expressions.

Thus, if five numbers had been required to be found, we should
have represented them by x — 2 2/, x — y, x, x + y, and x + 2 y.

3. Find the sum of all the integers beginning with 1
and ending with 100.

4. Find the sum of all the even integers beginning with
2 and ending with 1000.

5. The 8th term of an arithmetic progression is 10, and
the 14th term is — 14. Find the 23d term.

6. Find four numbers in arithmetic progression such
that the sum of the first two shall be 12, and the sum of
the last two — 20.

298 ALGEBRA.

7. Find the sum of the first 15 positive integers which
are multiples of 7.

8. The 19th term of an arithmetic progression is 9x—2y,
and the 31st term is 13x —8y. Find the sum of the first
thirteen terms.

9. Find four integers in arithmetic progression such that
their sum shall be 24, and their product 945.

10. How many positive integers of three digits are there
which are multiples of 9 ?

11. Find the sum of all positive integers of three digits
which are multiples of 11.

12. The 7th term of an arithmetic progression is — ^, the
16th term is ^, and the last term is J^. Find the number
of terms.

13. The sum of the 2d and 6th terms of an arithmetic
progression is — f, and the sum of the 5th and 9th terms is
— 10. Find the first term.

14. Find five numbers in arithmetic progression such
that the sum of the second, third, and fifth shall be 10, and
the product of the first and fourth — 36.

15. If m arithmetic means be inserted between a and b,
what is the first mean ?

16. How many positive integers of one, two, or three
digits are there which are multiples of 8 ?

17. How many arithmetic means are inserted between 4
and 36, when the second mean is to the first as 4 is to 3 ?

18. A man travels 3 miles the first day, 6 miles the
second day, 9 miles the third day, and so on. After he has
travelled a certain number of days, he finds his average
daily distance to be 46|- miles. How many days has he been
travelling ?

PROGRESSIONS. 299

19. How many arithmetic means are inserted between f
and —•^, when the sum of the first two is -^?

20. After A had travelled for 4i hours at the rate of 5
miles an hour, B set out to overtake him, and travelled 3
miles the first hour, 3|- miles the second hour, 4 miles the
third hour, and so on; in how many hours will B over-
take A ?

21. Find three numbers in arithmetic progression such
that the sum of their squares is 347, and one-half the third
number exceeds the sum of the first and second by 4i-.

22. The digits of a number of three figures are in arith-
metic progression ; the sum of the first two digits exceeds
the third by 3; and if 396 be added to the number, the
digits will be inverted. Find the number.

GEOMETRIC PROGRESSION.

342. A Geometric Progression is a series of terms each
of which is derived from the preceding by multiplying by a
constant quantity called the ratio.

Thus, 2, 6, 18, 54, 162, •••is a geometric progression in
which the ratio is 3.

Again, 9, 3, 1, -j, ^, • • • is a geometric progression in which
the ratio is ^.

Negative values of the ratio are also admissible.

Thus, — 3, 6, — 12, 24, — 48, • • • is a geometric progression
in which the ratio is — 2.

343. Given the first term, a, the ratio, r, and the number
of terms, n, to find the last term, I.

The progression is a, ar, ar^, ar^, •••.

It will be observed that the exponent of r in any term is
1 less than the number of the term.

Then in the nth or last term the exponent oi r is n — 1.
That is, I = ar"-\ (I.)

300 ALGEBRA.

344. Given the first term, a, the last term, I, and the ratio^
r, to find the sum of the terms, S.

S = a -{- ar + aif^ + ■" + ar"'^ + ar"~^ + ar""'^.

Multiplying each term by r, we have

rS = ar + ar^ + ai-^ 4- • • • + ar''^^ + ar'^~^ + ar".

Subtracting the first equation from the second,

rS — S = ar"" — a.

Whence, S = ^^" ~ ^.

r-1
But by (I.), § 343, rl = ar\

Therefore, S = ^i::!^. (II.)

r — 1

EXAMPLES.

345. 1. Find the last term and the sum of the terms of

the progression 3, 1, -, ••• to 7 terms,
o

In this case, a = 3, r = -, and n = 7.

Substituting in (I.), l^si^Y z=l = ~

243

lxJ--3 J--3 2186
o V. .-. .• • /TT N o 3 243 729 729 1093
Substituting in (II. ), S = — = — = — = -^^ •

3~^ ~3 "i

Note. The ratio may be found by dividing the second term by the
first, or any term by the next preceding term.

2. Find the last term and the sum of the terms of the
progression — 2, 6, — 18, ••• to 8 terms.

In this case, a = — 2, r = = — 3, and n = 8.

-2

Then, Z = - 2(- 3)^ = - 2 x (- 2187)= 4374.

And, ^^ -3 x4374-(-2)^- 1.3122 + 2^3^30^
-3-1 _4

PROGRESSIONS. 30]

Find the last term and the sum of the terms of :

3. 1, 3, 9, • • • to 8 terms.

4. 6, 4, f , ••• to 7 terms.

3

5. - 2, 10, - 50, ••• to 5 terms.

6. 2, 4, 8,..- to 11 terms.

3 3

7. — 3, -,—-,••• to 9 terms.

8. -|, -5, -10,..- to 10 terms.

9. — 5, 2, —-,••• to 6 terms.

5

10. --, -, --^'•' to 7 terms.

11. -^ 2' §'"' to 5 terms.

12. --, 3, - 12, ... to 6 terms.

346. If any three of the five elements of a geometric
progression are given, the other two may be found by sub-
stituting the given values in the fundamental formulae (I.)
and (II.), and solving the resulting equations.

But in certain cases the operation involves the solution
of an equation of a degree higher than the second ; and in
others the unknown quantity appears as an exponent, the
solution of which form of equation can usually only be
affected by the aid of logarithms (§ 419).

In all such cases in the present chapter, the equations
may be solved by inspection.

1. Given a = — 2, w = 5, Z = — 32 ; find r and S.
Substituting the given vahies in (I.), we have

— 32 = - 2 »•* ; whence, »•* = 16, and r = ± 2.

302 ALGEBRA.

Substituting in (II. )»

-2-1 -3

Therefore, r = 2 and S = - 62 ; or, ?■ = - 2 and S = -22, Ans.

Note 1. The interpretation of the two answers is as follows :

If r- = 2, the progression is — 2, — 4, — 8, — 16, — 32, whose sum

is - 62.

If r = — 2, the progression is — 2, 4, — 8, 16, — 32, whose sum is

-22.

2. Given a = 3, r = — -, S=^-—^; find n and I.

-lz-3
„ , . . . .xTx 1640 3 Z + 9

Substituting m (II. ) , -^^ = j = — ^ —

"3"

Whence, z + 9 =^560 6560 _g ^_ 1

729 729 729

Substituting the values of I, a, and r in (I.), we have

729 \ 3) \ 3) 2187

Whence, by inspection, ?j — 1 = 7, or n — 8.

EXAMPLES.

3. Given r = 2, n = 9, 1 = 256 ; find a and S.

4. Given ?• = -, n = 5, S = '^; find a and Z.

3 27 '

5. Given a = -2, n = 6, 1 = 2048 ; find r and S.

6. Given a = 2, ?• = , l = — —-; find n and aS.

2 256

7. Given r = i, n = ll, ^ = ^5 find a and I

2 204o

8. Given a = |, n = 9, l = ^f^; find r and .S.

3 12o

PROGRESSIONS. 303

9. Given a = — 8, 1 = — —, S = — — ; find r and n.

10. Given a = -, r = , 8=--—: find Z and n.

4 3 162'

11. Given 1 = 192, r = - 2, ^ = 129; find a and n.

2 1 '^^^

12. Given a = — ^, l = — zr:-rr, S = — ^^\ find r and ra.

3 192' 192'

From (I.) and (II.), general formnlae may be derived for
the solution of cases like the above.

13. Given a, r, and S ; derive the formula for I.

14. Given a, I, and S ; derive the formula for r.

15. Given r, I, and S ; derive the formula for a.

16. Given r, n, and I ; derive the formulae for a and S.

17. Given r, n, and S; derive the formulae for a and I.

18. Given a, n, and I ; derive the formulae for r and S.

Note 2. If the given elements are n, I, and ^S*, equations for a
and r may be found, but there are no definite formuloe for their
values. The same is the case when the given elements are a, n,
and S.

The general formulae for n involve logarithms ; these cases are
discussed in § 419.

347. The limit (§ 292) to which the sum of the terms of
a decreasing geometric progression approaches, when the
number of terms is indefinitely increased, is called the sum
of the series to infinity.

Formula (II.), § 344, may be written
CY a — rl
1 — r

It is evident that, by sufficiently continuing a decreasing
geometric progression, tlie last term may be made numeri-
cally less than any assigned number, however small.

Hence, when the number of terms is indefinitely increased,
I, and therefore rl, approaches the limit 0.

304 ALGEBRA.

Then the fraction ^~ ^ approaches the limit

1 — r 1 — r

Therefore, the sum of a decreasing geometric progressioE
to infinity is given by the formula

s

1-

• r

(III.)

EXAMPLES

1. Find the sum of the

series

4, -

8 16

3' 9' ■"'

to infinity.

2
lu this case, a — i, ?• = — -•

o

Substituting in (III.), ^ = -

4

12

Ans.

Find the sum of the followini

g to

infinity :

2. 3, 1, |, ....

6.

7 21 63
4' 32' 256'

....

3. 16, -4,1, •...

7.

2 15
5' 3' 18'

.-..

4 _1 1 -J. ...

' 5' 25' •

8.

1 1

8' 18'

2
""81'*'*'

. 5 10 20

^- 3' 9' 27'*"'

9.

5 5 35

7' 8' 64'

....

348. Tofiyid the value of a repeating decimal.

This is a case of finding the sum of a decreasing geometric
series to infinity, and may be solved by formula (HI-)-

1. Find' the value of .85151 ••..

We have, .85151 .•• = .8 + .051 + .00051 + •••.

The terms after the first constitute a decreasing geometric pro

gression, in which a = .051 and r — .01.

cj V *•* ♦• • /TTT> c -051 .051 51 17
Substitutmg m (III.), S = ^— ^ = "1^ = ^ = 33-0'

8 17 281

Then the value of the given decimal is h -— , or -— , Ans.

10 oo\j ooU

PROGRESSIONS. 305

EXAMPLES.
Find the values of the following :

2. .8181.... 4. .69444.... 6. .11567567....

3. .296296.... 5. .58686.... 7. .922828.-..

349. To insert any number of geometric means between two

given terms.

128

1. Insert 5 geometric means between 2 and

729

We are to find a geometric progression of 7 terms, whose first

term is 2, and last term

' 729

128
Putting a = 2, I- ^, and n — 7, in (I.), § 343, we have

128 „. , ,64 , ,2

— = 2 J* ; whence, r' — — , and r = ± -•
729 729 3

Hence, the required result is

2, ±4 8 16 3^ 61 128 ^^^^ ^

3 9 27 81 243 729

EXAMPLES.

2. Insert 4 geometric means between 3 and 729.

3. Insert 6 geometric means between - and .

^ 6 3

4. Insert 5 geometric means between 2 and 128.

2 125

5. Insert 3 geometric means oetween — - and -—•

5 8

7

6. Insert 4 geometric means between — - and 3584.

LI

243 2

7. Insert 7 geometric means between — — and --.

128 27

350. Let X denote the geometric mean between a and &.

Then, by the nature of the progression, - == -, or ar^ = ah.

ax

306 ALGEBRA.

Whence, x = Va6.

That is, the geometric mean between two quantities is equal
to the square root of their product.

EXAMPLES.

Find the geometric mean between :

1. 2iiandli|. 2. 9 + 4 VS and 9 - 4 VB.

3. a^ + 2 a6 + 6- and a^ - 2 ab + b\

2x^ + 4a;y ^y + '^y-

xy — 2y'^ 2x^~4:xy

PROBLEMS.

351. 1. Find three nnmbers in geometric progression such
that their sum shall be 14, and the sum of their squares 84.

Let the numbers be a, ar, and ar'^.

a + ar + ar^ = 14. (1)

Then by the conditions, ,

' a2 + «->••- + cih-^ = 84. (2)

Dividing (2) by (1), a - ar + af^ = 6. (3)

Subtracting (3) from (1), 2 ar —8, or r = — (4)

a

Substituting in (1), a + 4 + — = 14, or a~ -10a = - 16.
a

Solving this equation, a = 8 or 2.

4 4 1

Substituting in (4), r = - or - = - or 2.

" ^ 8 2 2

Therefore, the numbers are 2, 4, and 8, Ans.

2. The 4th tenn of a geometric progression is — ^, and
the 7th term is Iff. Find the second term.

3. The sum of the first and last of four numbers in geo-
metric progression is 112, and the sum of the siecond and
third is 48. Find the numbers.

4. The product of three numbers in geometric progres-
sion is — 1000, and the sum of the squares of the second
and third is 500. Find the numbers.

PROGRESSIONS. 307

5. A man saves every year half as much again as he
saved the preceding year. If he saved $ 128 the first year,
to what sum will his savings amount at the end of seven
years ?

6. A body moves 12 feet the first second, and in each
succeeding second five-eighths as far as in the preceding
second, until it comes to rest. How far will it have moved '!

7. The 5th term of a geometric progression is — f, and
the 9th term is - i^f. Find the 11th term.

8. If m geometric means be inserted between a and h,
what is the first mean ?

9. The sum of three numbers in arithmetic progression
is 12. If the first number be increased by 5, the second by
2, and the third by 7, the resulting numbers form a geo-
metric progression. What are the numbers ?

10. Divide $ 700 betAveen A, B, C, and D, so that their
shares may be in geometric progression, and the sum of A's
and B's shares equal to $ 252.

11. There are four numbers, the first three of which form
an arithmetic progression, and the last three a geometric
progression. The sum of the first and third is 2, and of
the second and fourth 37. What are the numbers ?

12. Find the ratio of the geometric progression in which
the sum of the first ten terms is 244 times the sum of the
first five terms.

13. There are three numbers in geometric progression
whose sum is 19. If the first be multiplied by |, the second
by I, and the third by |, the resulting numbers form an
arithmetic progression. What are the numbers ?

HARMONIC PROGRESSION.

352. A Harmonic Progression is a series of terms whose
reciprocals form an arithmetic progression.

308 ALGEBRA.

Thus, 1, ^, i, f, i, ••• is a harmonic progression, because
the reciprocals of the terms, 1, 3, 5, 7, 9, •••, form an arith-
metic progression.

353. Any problem in harmonic progression which is sus-
ceptible of solution, may be solved by taking the reciprocals
of the terms, and applying the formulae of the arithmetic
progression. There is, however, no general method for
finding the sum of the terms of a harmonic progression.

354. Let X denote the harmonic mean between a and h.

Then, - is the arithmetic mean between - and j- (§ 352).
X a 0 ■'

Whence by § 340, - = — ^7— = ^ ^ , and x = r •

■^ ' X 2 2ab ' a + b

BXAMPLES.

355. 1. Find the last term of the progression 2, |, f, •••
to 36 terms.

Taking the reciprocals of the terms, we have the arithmetic pro-

1 3 5
gression ■. -> -» •••.

In this case, a —-, d = I, and n = 36.

2

Substituting in (I.), § 334, we have I = } + (36 - 1) x 1 = — •
Taking the reciprocal of this, the last term of the given harmonic
progression is — , Ans.

2. Insert 5 harmonic means between 2 and — 3.

We have to insert 5 arithmetic means between - and — •

2 3

Putting a = -, I =— , and n = 7, in (I.), § 334, we have
2 3

11 5 5

— = - + 6 d ; whence, 6d = , or d =

3 2 6 36

PROGRESSIONS. 309

Then the arithmetic progression is

1 13 2 _1^ _J_ _T_ _1
2' 36' 9' 12' 18' 36' S

Therefore, the required harmonic progression is

2, ^, ?, 12, -18, -^, -3, Ans.
' 13 2 7

Find the last terms of the following :

3. -, -, 1, ••• to 13 terms. 4. -, — -, — r, ••• to 25 terms

5 7 5 43 71

5. -3, 2, ?, ... to 38 terms.
4

a 4 6 12 . .Q,

6. , , , ... to 43 terms.

3' 5' 11'

5 2 5

7. , — -, , ••• to 17 terms.

6' 3' 9

8. Insert 6 harmonic means between 2 and

9

2 2

9. Insert 7 harmonic means between and -•

5 7

10. Insert 8 harmonic means between and

5 5

Find the harmonic mean between :

11. 3 and 6. 12. ^-^ and - ^—^^

1 + x 1 +a^

13. The first term of a harmonic progression is x, and the
second term is y ; continue the series to three more terms.

14. The arithmetic mean between two numbers is 1, and
the harmonic mean — 15. Find the numbers.

15. The 5th term of a harmonic progression is — ^, and
the 11th term is — ^. What is the 15th term ?

16. Prove that, if a, b, and c are in harmonic progression,

a: c = a — b :b — c.

310 ALGEBRA.

XXXI. THE BINOMIAL THEOREM.

POSITIVE INTEGRAL EXPONENT.

356. The Binomial Theorem is a formula by means oi
which any power of a binomial, positive or negative, inte-
gral or fractional, may be expanded into a series.

We shall consider in the present chapter those cases
only in which the exponent is a positive integer.

357. Proof of the Binomial Theorem for a Positive Inte
gral Exponent.

By actual multiplication, we obtain :

(a -\- xy = a^ -{- 2 ax + a^ ;

(a + xf = a^ + 3 a^x + 3 ax? + a?',

(a + xy = a* + 4. a^x + 6 a^x'^ + 4 aa^ + a;*; etc.

In the above results, we observe the following laws :

1. The number of terms is greater by 1 than the expo-
nent of the binomial.

2. The exponent of a in the first term is the same as the
exponent of the ' binomial, and decreases by 1 in each suc-
ceeding term.

3. The exponent of x in the second term is 1, and in-
creases by 1 in each succeeding term.

4. The coefficient of the first term is 1, and the coefficient
of the second term is the exponent of the binomial.

5. If the coefficient of any term be multiplied by the
exponent of a in that term, and the result divided by the
exponent of x in the term increased by 1, the quotient will
be the coefficient of the next following term.

THE BINOMIAL THEOREM. 311

358. If the laws of § 357 be assumed to hold for the
expansion of (a + •^)"> where n is any positive integer, the
exponent of a in the first term is 7i, in the second term
w — 1, in the third term n — 2, in the fourth term n — 3, etc.

The exponent of x in the second term is 1, in the third
term 2, in the fourth term 3, etc.

The coefficient of the first term is 1 ; of the second term n.

Multiplying the coefficient of the second term, n, by n—1,
the exponent of a in that term, and dividing the result by
the exponent of x in the term increased by 1, or 2, we have

—^ — — ^ as the coefficient of the third term ; and so on.
Then, (a + xf = a» + 7ia»-^ x + '^^^^-J-) al'-^n?

^ni!!L=m^a-'-V+.... (1)

Multiplying both members of (1) by a + a;, we have

(a + 0;)"+^ = a"+i + nO'x + '^i^-^) a»- V

n{n-l){n-2) 2 «
^ 1.2.3 '^ "^^

+ aPx + wa"-V + ^ ^^ ~ -'-^ a"-2a^ + ... .

Collecting the terms which contain like powers of a and a;,

(a + a;)«+^ = a"+^ + (n + 1) a"a; + ["^(^-1) 4. ^1 a^-^a?

rn(n - l)(n -2) n{n-l) \ ,^ ^_
L 1.2.3 ^ 1-2 J ^

4.?L(^r!^ + lJan-.^^.....

* A point is often used in place of the sign x ; thus, 1 • 2 is the
same as 1 x 2.

312 ALGEBRA.

Or, (a + «)"+' = a"+^ + (w + 1) a"a; + n f^^^^^l a^-^ic"

, w (w — 1) fw + in __2 , ,

= «"' +(m + 1) a'x + ^"+^)'' a"-W

JL • ^

, (n + l)n(n — l) „_2 « ,
^^ ^ "'

It 'will be observed that this result is in accordance with
the laws of § 357 ; which proves that, if the laws of § 357
hold for any power of a + a; whose exponent is a positive
integer, they also hold for a power whose exponent is
greater by 1.

But the laws have been shown to hold for (a + x)*, and
hence they also hold for (a + xy ; and since they hold for
(a + xy, they also hold for (a + xy ; and so on.

Therefore, the laws hold when the exponent is any posi-
tive integer, and equation (1) is proved for every positive
integral value of n.

Equation (1) is called the Binomial Theorem.

Note 1. The above method of proof is known as Mathematical
Induction.

Note 2. In place of the denomhiators 1.2, 1-2 -3, etc., it is
usual to write [2, [3, etc. The symbol \n, read '■'■factorial n," signifies
the product of the natural numbers from 1 to m inclusive.

359. Putting a = 1 in equation (1), § 358, we have

(1 +a.)» = l +na.4-^%:^a^+^^*^-^^^a^+ ....

EXAMPLES.

360. In expanding expressions by the Binomial Theorem,
it is convenient to obtain the exponents and coefficients of
the terms by aid of the laws of § 357, which have been
proved to hold for any positive integral exponent.

THE BINOMIAL THEOREM. 313

1. Expand (a + xy.

The exponent of a in the first term is 5, in the second term 4, in
the third term 8, in tlie fourth term 2, in tlie fiftli terra 1.

The exponent of x in the second term is 1, in the third term 2, in
the fourtli terra 3, in the fifth term 4, in the sixth term 5.

The coefficient of the first term is 1 ; of the second term, 5.

Multiplying the coefficient of the second term, 5, by 4, the exponent
of a in that term, and dividing the result by the exponent of x in the
term increased by 1, or 2, we have 10 as the coefficient of the third
term; and so on.

Then, (« + x)^ = a^ + 5 a*x + 10 a^x^ + 10 a'^x^ + 5 ax* + x^, Ans.

Note 1. The coefficients of terms equally distant from the begin-
ning and end of the expansion are equal. Thus the coefficients of the
latter half of an expansion may be written out from the first half.

If the second term of the binomial is negative, it should
be enclosed, sign and all, in a parenthesis before applying
the laws. In reducing afterwards, care must be taken to
apply the principles of § 186.

2. Expand (1 - xy.

We have, (l-x)6 = [l + (-x)]8

= 16+6.15.(-x) + 15.1*.(-x)2+20.18.(-x)*

+ 15 . 12 . (_ X)4 + 6 . 1 . (- X)5 + (- X)6

= 1 - 6 X + 15x2 - 20 X* + 15 X* - 6 x^ + x^, Ans.

Note 2. If the first term of the binomial is numerical, it is con-
venient to write the exponents at first without reduction. The result
should afterwards be reduced to its simplest form.

If either term of the binomial has a coefficient or exponent
other than unity, it should be enclosed in a parenthesis be-
fore applying the laws.

3. Expand {3m^-</7iy.

(3 m2 - v^)4 = [(3 m2) + ( - n*)]4

= (3m2)* + 4(3m2)3(- n^)+ e(3m^)^(- n^)^

+ 4(3m2)(-n*)3 + (_n^)4
= 81 to8 _ 108 r?i«K^ + 54 m*n^ - 12 mH + nt Ans.

314

ALGEBRA.

E^s

.pand the following:

4.
6.

[x + iy.

[a + xf.

15. (1+2 my.

16. (1-xy.

24.

(-*a'

6.

7.
8.

[a - xy.

(m — ny.

[1 + xy.

17. '(</x' + x-y.

18. (a* -2/.

19. (3 + a^y.

25.
26.

(2a-'3-V&)^

9.
10.

(a - by.

(a- + b'cy.

20. ('i, + ^'.

27.

(V*-^^J-

11.

[x^"* + f"y.

21. (4.ai-x^y.

28.

(Va^ + 4-^a)*.

12.
13.

[2a-iy.

(x + 2y.

n. (.^-0.

29.

14.

[a -3 by.

23. {m^-{-5x-y.

30.

(2 a - 3 by.

31. (ah~i + (

i^h^y. 32.

V ^ n ^m

A trinomial may be raised to any power by the Binomial
Theorem if two of its terms be enclosed in a parenthesis
and regarded as a single term.

33. Expand (x--2x- 2y.

(x^-2x-2y =[(x2-2x) + (-2)]*

= (x2 - 2 x)* + 4(x2 - 2 xy(- 2) + 6(«2 - 2 x)2(- 2)«

+ 4(x2-2x)(-2)3 + (-2)4
= x8-8x''+24x6-32x5+16x*-8(x6-6x5+ 12x4-8x3)

+ 24(x* - 4 x3 + 4 x2) - 32(x2 _ 2 x) + 16
= x8_8xH16x6 + 16x5-56x4-32x3
+ 64 x2 + 64 X + 16, Ans.

Expand the following :

34. (l-x + c^\ 37. (:?r'-2a;-3)*

. 35. (.^•- + a; + 2)*. 38. (1 + a; - x'y.

36. (l + 3a;-ccy. 39. (a^ - a; + 2/.

THE BINOMIAL TIIEOKEM. 315

361. To find the rth or general term in the expansion of
(a + xy\

The following laws will be found to hold for any term
in the expansion of (a + x)", in equation (1), § 358 :

1. The exponent of x is less by 1 than the number of
the term.

2. The exponent of a is n minus the exponent of x.

3. The last factor of the numerator is greater by 1 than
the exponent of a.

4. The last factor of the denominator is the same. as the
exponent of x.

Hence, in the ?'th term, the exponent of a; is r — 1.
The exponent of a is w — {r — 1), or n — r + 1.
The last factor of the numerator is n — ?' + 2.
The last factor of the denominator is r — 1.
Therefore, the ^-th term of the expansion of (a + xy is
w (n - 1) (n - 2) • • • (n - r + 2) ^„_,+i^,-i
1.2.3..- (r-1)

EXAMPLES.

362. In finding any term of an expansion, it is convenient
to obtain the coefficient and the exponents of the terms by
aid of the laws of § 361.

1. Find the 8th term of (3a^ - h-y\

We have, (3a^ - b~'^y^ = [(3a^) + (- ft-i)]".

In this case, n — \\ and r — 8.

The exponent of (— 6-i) is 8 — 1, or 7.

1
The exponent of (3a^) is 11 — 7, or 4.

The first factor of the numerator is 11, and the last factor 4 + 1,
or 5.

The last factor of the denominator is 7.'

Hence, the 8th term ^ 1^ • ^0 • 9 • 8 • 7 • 6 • 5 .3 ^^^^^ ^_^y
1.2.3.4.5.6.7 ^ ^ ^

= 330 • (81 a2)(- 6--) = -26730a26-^ Ans.

316

ALGEBRA.

Note. If the second term of the binomial is negative, it should be
enclosed, sign and all, in a parenthesis before applying the laws.

If either term of the binomial has a coefficient or exponent other
than unity, it should be enclosed in a parenthesis before applying the
laws.

Find the

2. 4th term of (a + xy.

3. 9th term of (m + 1)".

4. 5th term of (a - bf.

5. 10th term of (1 - xf^

6. 9th term of (m^ — n^y^.

7. 7th term of fa-' + ^Y"

8. 4th term of (~ - ^Y'-

9. 10th term of (a" + a")'"-
10. 8th term of (x-^ -2y^y\

11. 6th term of (a"^ + 3ar*)'«

12. 7th term of f a^ ^^ ^

4V6y

13. 8th term of (cC^ + 2-Vx'y\

14. 5th term of (m^ + -

9 \15

15. 6th term of (x-^-^^^ •

16. Find the middle term of ( a^ - -^ *«

UNDETERMINED COEFFICIENTS. 317

XXXII. UNDETERMINED COEFFICIENTS.

CONVERGENCY AND DIVERGENCY OF SERIES.

363. A Series is a succession of terms so related that
e;ich may be derived from one or more of the preceding in
accordance with some tixed law.

A Finite Series is one having a finite number of terms.

An Infinite Series is one the number of whose terms is
unlimited.

The progressions, in general, are examples of finite series,
but in § 347 we considered infinite geometrical series.

364. Infinite series may be developed by Division.
Let it be required, for example, to divide Ihj 1 — x.

l-x)l(l + x + x-+ ...
1-x

X

X — x^

Therefore, —l-{-x + x^-\--".

1 — x

Infinite series may also be developed by Evolution (see
Exs. 25 to 30, § 195), and by other methods, one of the most
important of which will be considered in § 369.

365. A series is said to be convergent either when the
sum of the first n terms approaches a certain fixed quantity
as a limit (§ 292) ; when n is indefinitely increased ; or when
the sum of all the terms is equal to a fixed finite quantity.

A series is said to be divergent when the sum of the first
n terms can be made to numerically exceed any assigned
quantity, however great, by taking n su^fiiciently great.

318 ALGEBRA.

366. Consider, for example, the infinite series

l-{-x-\-x- + x^-\ .

I. Suppose X = aji, wliere x^ is numerically < 1.
The sum of the first n terms is now

1 + xi + ctf + ... + xr' = ^^=^ (§ 86).

1 — Xi

If n is indefinitely increased, Xi' decreases indefinitely in

absolute value, and approaches the limit 0.

1 — a; " .1

Then the fraction "-^ approaches the limit

X — Ct/j J- Xj

That is, the sum of the first n terms approaches a certain
fixed quantity as a limit, when n is indefinitely increased.
Hence, the series is convergent Avhen x is numerically < 1.

II. Suppose x = l.

In this case, each term of the series is equal to 1, and the
sum of the first n terms is equal to n ; and this sum can be
made to exceed any assigned quantity, however great, by
taking 7i sufficiently great.

Hence, the series is divergent when x = l.

III. Suppose a; = — 1.

In this case, the series takes the form 1 — 1 + 1 — 1 + .-.,
and the sum of the first n terms is either 1 or 0 according
as n is odd or even.

Hence, the series is neither convergent nor divergent when
x = -l.

IV. Suppose x = Xi, where x^ is numerically > 1.
The sum of the first n terms is now

1 + 0^1 + a;/ + ... + xr' = ^^-=^ (§ 86).

Xi — 1

By taking n sufficiently great, — ' — can be made to

numerically exceed any assigned quantity, however great.

UNDETERMINED COEFFICIENTS. 319

Hence, the series is divergent when x is numerically > 1.

367. Consider the infinite series

l + a;-|-ar + .^+...,

developed by the fraction (§ 364),

Let X = .1, in which case the series is convergent (§ 366).
The series now takes the form 1 + .1 + .01 + .001 + •••,

while the value of the fraction is — , or —

.9' 9

In this case, however great the number of terms taken,

their sum will never exactly equal — ; but it approaches
this value as a limit. (See § 347.)

Thus, if an infinite series is convergent, the greater the
number of terms taken, the more nearly does their sura
approach to the value of the expression from which the
series was developed.

Again, let x = 10, in which case the series is divergent.

The series now takes the form 1 + 10 + 100 + 1000 -\ ,

while the value of the fraction is , or

1-10' 9

In this case it is evident that, the greater the number of
terms taken, the more does their sum diverge from the

value

9

Thus, if an infinite series is divergent, the greater the
number of terms taken, the more does their sum diverge
from the value of the expression from which the series was
developed.

It follows from the above that an infinite series cannot be
used for the jyurposes of demonstration, unless it is coyivergent.

368. The infinite series

a + hx -\- cx^ + dz^ + • • •

is convergent when x = 0 ; for the sum of all the terms is
equal to a when a; = 0.

320 ALGEBRA.

THE THEOREM OF UNDETERMINED COEFFICIENTS.

369. An important method for expanding expressiona

into series is based on the following theorem :

The Theorem of Undetermined Coeflficients.

If the series A + Bx + Ox- + Dx^ + • • • is always equal to
the series A' + B'x -{- C'x' + i)'x^ + •••, when x has any value
ivhich makes both series convergent, the coefficients of like
power's of X in the series will be equal; that is, A — A',
B=^B', C^ C, etc.

For since the equation

A + Bx + Ca^ + Djc^ + — = A' -h B'x -^ C'x"" + D'a^ + —

is satisfied when x has any value which makes both mem-
bers convergent, and since both members are convergent
when a; = 0 (§ 368), it follows that the equation is satisfied
when x = 0.

Putting a: = 0, we have A = A'.

Subtracting A from the first member of the equation, and
its equal A' from the second member, we obtain

Bx + Cx^ + I)x'+ ■■• = B'x + C'x^ + D'x? + ••••

Dividing each term by x,

B-\-Cx + Dx'+--- = B' -\- C'x + D'y? + ....

This equation also is satisfied when x has any value
which makes both members convergent ; and putting ic = 0,
we have B = B'.

In like manner, we may prove C = C, D = D', etc.

370. A finite series being always convergent, it follows
from the preceding article that if two finite series

A + Bx+Cx--\ hAx" and A' + B'x-\-C'x--\ hK'x''

are equal for every value of x, the coefficients of like powers
of X in the two series are equal.

UNDETERMINED COEFFICIENTS. 321

EXPANSION OF FRACTIONS INTO SERIES.

2 — »•? c^ — v^

371. 1. Expand ' '-— in ascending powers of x.

1 — 2 X + 'Sx^

Assume 2 -3x^-x^ = A + Bx + Cx? + Dx^ + j&x* + ... : (11
1 - 2 a; + 3 x'^ » v >

where A, B, C, D, E, etc., are quantities independent of x.

Clearing of fractious, and collecting the terms in the second mem-
ber involving like powers of x, we have

2-3x:^-x^ = A+ B x+ C x2+ D x^+ E x* + .... (2)
-2A -2B -2C -2D
+ SA +3B +'SC

The second member of (1) must express the value of the fraction
for every value of x which makes the series convergent (§ 367).

Hence, equation (2) is satisfied when x has any value which makes
both members convergent ; and by the Theorem of Undetermined
Coetiicients, the coefficients of like powers of x in the series are equal.

Then, A = 2.

B-2A = 0; whence, B-2A =4.

C_2B + 3^ = -3; whence, C = 2B-3^-3 = -l.

j)-20 + SB = -l; whence, Z) = 2 C - 3 J5 - 1 = - 15.

E-2D + 3C = 0; vfhence, E - 2 D - S C = - 27 ; etc.
Substituting these values in (1), we have

2-3x2-x« =2 + 4x-x2-15x3-27x^+ •••, Ans.
1 - 2 X + 3 x2

The result may be verified by division.

Note 1. A vertical line, called a bar, is often used in place of a
parenthesis.

Thus, + B \ X is equivalent to (i? — 2 A'jx.

-2a\

Note 2. The result expresses the value of the given fraction only
for such values of x as make the series convergent (§ 367).

If the numerator and denominator contain only even
powers of x, the operation may be abridged by assuming a
series containing only the even powers of x.

322 ALGEBRA.

2 -I- 4,x^ — X*

Thus, if the fraction were — — , we should as-

' l-Sx' + Bx*'

sume it equal to J. + Bx^ + Cx* + Dx^ + iV + • • •.

In like manner, if the numerator contains only odd
powers of x, and the denominator only even powers, we
should assume a series containing only the odd powers of x.

If every term of the numerator contains x, we may as-
sume a series commencing with the lowest power of x in
the numerator.

If every term of the denominator contains x, we determine
by actual division what power of x will occur in the first
term of the expansion, and then assume the fraction equal
to a series commencing Avith this power of x, the exponents
of x in the succeeding terms increasing by unity as before.

2. Expand — in ascending powers of x.

Sar — or

/V.-2

Dividing 1 by 3 x-, the quotient is — ; we then assume

^ ^x-2 + Sc-i + C+ Dx + Ex^ + •••. (1)

3 z^ - x3
Clearing of fractions, we have

1 = 3 A + 3 B \ X + 3 C \ x"^ + 3 D \ x^ + 3 E \ z* + •-.
-^I-^I - C\ -2)1

Equating the coefficients of like powers of a;,

3^ = 1.

3B-A = 0.

SC-B = 0.

3D- C = 0.

3E -D = 0; etc.

Whence, A=\ -B = -, 0 = —, D = —, E = ~, etc.

3 9 27 81 243

Substituting in (1), we have

1 /V.-2 /J.-1 1 /v. ~2

3x2 -a8 3 9 ' 27 H\ 243

UNDETERMINED COEFFICIENTS.

323

EXAMPLES.

Expand each of the following to five terms in ascending
powers of x :

2x + 3x^-a^ ,o l-7a^-4x3

3.

6.

7.

1 +5x

1 + x

3-2a;

l-4aj

2 + 7a!2

l-3a^

4 a; — a^

2 + Sx'

1-x-

3x2

8.

9.

l_2a;-ar^

10.

11.

12.

l + 5x
1

2x'

6x'

-2x

2-3x + 4:X^

l-4a^ + 6ar^

1 + 2a;-a;2 *

2 + a; - 3 a;-

4a; + Sx^

13.

14.

15.

16.

17.

a^-5x*-2o^
3 + 5a;-2a;^
a^-3a^-\-x*'
x^ — 4:0!^ -{- 2x^
2-3a^-a^ '

2-3a;^
3 - 2 a; + a;3'

3 - 4 a.-^

2 a; + af^ - 3 a;*'

EXPANSION OF RADICALS INTO SERIES.

372. 1. Expand Vl — a; in ascending powers of x.
Assume Vl - x =^ A + Bx + Cx"^ + Dx^ + Ex* + •■■.
Squaring both members, we have by the rule of § 187,

1 - a; = ^2

+ 2AB

X + B^
+ 2AC

+ 2 AD
-\-2BC
Equating the coefficients of like powers of x,
A^ = \; whence, A=\.
2AB =

X'

+ C2

+ 2AE

+ 2BD

x*+-

1 ; whence, B = —

2A 2

B^ + 2AC = 0;

2AD + 2BC=0;

C'^ + 2AE + 2BB = 0;
Substituting these valut
Vl - X = 1 -
The result may be verified by evolution.

732 1

whence, C = =

2A 8

T5/-Y 1

whence, D = • =:

A 16

, „ C^ + 2BB

whence, E = -^— ; =

2 A

_ 5

128

in (1), we have

?_^'_^_5^_..., ^„s.

2 8 16 128

(1)

etc.

324 ALGEBRA.

EXAMPLES.

Expand each of the following to five terms in ascending
powers of x :

2. Vl+4a;. 4. VI + 2 a; - a^. 6. "v/l + 3 a;.

3. Vl-5a;. 5. VI - a; - a^. 7. ^l-a; + a^.

PARTIAL FRACTIONS.

373. If the denominator of a fraction can be resolved
into factors, each of the first degree in x, and the numerator
is of a lower degree than the denominator, the Theorem of
Undetermined Coefficients enables us to express the given
fraction as the sum of two or more partial fractions, whose
denominators are factors of the given denominator, and
whose numerators are independent of x.

374. Case I. When no two factors of the denominator
are equal.

1. Separate — '-^ into partial fractions.

(3a;-l)(5a; + 2)

Assume ^^^ + 1 =.^L_ + _^_, (i)

(3x-l)(5x + 2) 3x-l 5x + 2 ^^

where A and B are quantities independent of x.

Clearing of fractions, 19 x + 1 = .4(5 x + 2) + ^(3 x - 1).

Or, 19x+ 1 =(5^ + 3B)x + 2^-^. (2)

The second member of (1) must express the value of the given frac-
tion for every value of x.

Hence, equation (2) is satisfied by every value of x ; and by § 370,
the coefficients of like powers of x in the two members are equal.

That is, 5 ^ + 3 _B = 19,

and 2 A- B=\.

Solving these equations, we obtain A = 2 and 5 = 3.

Substituting in (1), ■ ^^^"'"^ = — ? — + — - — , Ans.

^ (3x-l)(5x + 2) 3x-l 5x + 2

The result may be verified by adding the partial fractions.

UNDETERMINED COEFFICIENTS. 325

2. Separate — ^^ — — into partial fractions.

The factors ot 2x — z^ — x^ are x, I — x, and 2 + x (§ 284).
Assume then ?-±^ = - + -^ + ^

25c-x2-x3 X 1-x 2 + x
Clearing of fractions, we have

X + 4 = ^(1 - X) (2 + x) + J5x(2 + x) + Cx(l - x).

This equation is satisfied by every value of x ; it is therefore satis-
fied when X = 0.

Putting X = 0, we have 4 = 2 A, or A = 2.

Again, the equation is satisfied when x = 1.

5
Putting X = 1, we have 6 = S B, or B = -•

The equation is also satisfied when x = — 2.
Putting X = — 2, we have 2 = — 6 C, or C = •

o

5 1

ThPn a; + 4 _2 3 .3

' 2x-x2-x3~x"^l-x^2 + x

^2 5 1 ^^^

X 3(1 -x) 3(2 + x)'

Note. To find the value of A, in Ex. 2, we give to x such a value
as will make the coefficients of B and C equal to zero ; and we proceed
in a similar manner to find the values of B and C.

This method of finding A, B, and C is usually shorter than that
used in Ex. 1.

EXAIVIPLES.
Separate each of the following into partial fractions :

q 18 a; 4- 3 c x^— 75 » ax—19a^

4ar^ — 9 x'^ — 25x cc^+4ax— 5a^

^ x-2 c 38a; + 5 q 46 -5a;

6x^-6x 6x' + 5x-6 8-18a;-5x2

9 x' + lOx-T jQ - 13 a;^ + 27x4-18

(2a;-l)(12a;2_a,_6y ' (a;^ _ 2 a;) (a;^ _ 9) *

326 ALGEBRA.

375. Case II. When all the factors of the denominator
are equal.

Let it be required to separate — ^ ^t — into partial

fractions. v ~ ^

Substituting y + 3 for x, the fraction becomes

(y + 3y-ll(y + 3) + 26_y'-5y + 2_l 5 ^ 2

Keplacing yhy x — 3, the result takes the form
1 5 2

a; - 3 (x-Sy {x- 3f

This shows that the given fraction can be expressed as
the sum of three partial fractions, whose numerators are
independent of x, and whose denominators are the powers
of a; — 3 beginning with the first and ending with the third.

Similar considerations hold with respect to any example
under Case II. ; the number of partial fractions in any case
being the same as the number of equal factors in the
denominator of the given fraction.

EXAMPLES.

Gaj 4- 5
376. 1. Separate— — -^-— into partial fractions.
( o X -\- oj

In accordance with the principle stated in § 376, we assume the
given fraction equal to the sum of tivo partial fractions, whose denomi-
nators are the powers of 3x + 5 beginning with the first and ending
with the second.

Thus, 6x + 5 ^^A_ B

(3x-f5)2 3x + 5 (3x + 5)2

Clearing of fractions, 6x + 5 = ^(3 x + 5) + B.

Or, 6x + 5 = 3^x + 5^ + 5.

Equating the coefficients of like powers of x,

3^ =6.

bA + B = b.

UNDETERMINED COEFFICIENTS. 327

Solving these equations, we have A = 2 and B = — 5.

Whence, 6a: + 5 ^ _2 5 ^^^^_

(3x + 5)2 3x + 6 (3x + 5)2'

Separate each of the following into partial fractions:

2 14 a; - 30 ^ 9a;^-15a;-l g 10 ar + 3 a; - 1
4ar-12a; + 0" ' {3x-iy ' ' (5x + 2y

3 x' + ^x-l g 8 ar- 19 - y?-3x^-x

(a; + 5)3 * ■ (2a;- 3/ ' (a;-l)*

g a;3 + 4a.'^ + 7a; + 2 g 18ar^- 21a^ + 4a;

(a; + 2/ * ■ (3a;-2y

377. Case III. When some of the factors of the denomi-
nator are equal.

1. Separate ^^—- into partial fractions.

^ x(x-{- ly

The method in Case III. is a combination of those of Cases I. and II.

^ x^-4x + SA^B^ C
We assume —- = — i +

X(X+1)2 X X+1 (iC + l)2

. Clearmg of fractions, x^ - 4x + 3 = A(x + 1)2 + Bx(x + 1) + Cx

= (A + Byx'^ + (2A + B+C)x + A.

Equating the coefficients of like powers of x,

A + B = l.

2A + B+ C = -4.

A = S.

Solving these equations, we have ^ = 3, B — — 2, and C = — 8.

Whence, x^ - 4x f 3 ^ 3 _ _2__ _8 ^^^

X(X +1)2 X X 4- 1 (X + 1)2

Note. It is impracticable to give an illustrative example for every
possible case ; but no difficulty will be found in assuming the proper
partial fractions if attention is given to the following general rule.

328 ALGEBRA.

ir

The fraction — should be put equal to

(« + a)(x + b) ••• (X + my ■■•

x + a x + b x + m (x + my {x + m)'

Single factors like x + a and x + b having single partial fractions
corresponding, arranged as in Case I. ; and repeated factors like
(x + my having r partial fractions corresponding, arranged as in
Case II.

EXAMPLES.
Separate each of the following into partial fractions :
18-5a;-3a;^ g 2 - 3a; - ar' - 2a^

Sx' + Sx'-lSx-S g 4.-9x-12x'-2a?

x* + 4.a^ ' ' x(x + l)(x-^2y

„ 12 a.-2 - 11 a; - 38 .^ 3x4-13

■ (3a;-l)(2a; + 3)2 (2 a;- 3)(8ar^ - lOx - 3)

378. If the degree of the numerator is equal to, or greater
than, that of the denominator, the preceding methods are
inapplicable.

In such a case, we divide the numerator by the denomi-
nator until a remainder is obtained which is of a lower
degree than the denominator.

1. Separate into an integral expression and

xi^ — x

partial fractions.

Dividing x^ — 3 x^ _ i by x^ — x, the quotient is x — 2, and the
remainder — 2x — 1.

Then, ^ ~ i = x - 2 + -^ -•

X2 — X X''^ — X

Separating ~ into partial fractions by the method of Case

X2 — X

I,, the result is —

X X — 1

Whence, - — ^ t = x-2 + - ^, Ans.

x2 - X X X - 1

UNDETERMINED COEFFICIENTS. 329

EXAMPLES.

Separate each of the following into an integral expression
and two or more partial fractions :

(a; + 2)(3a)-l) x'(x + l)

„ 2a^-nx' + Ux-29 g a^-2r' + 4a;-l

{x-2y ' ' x\x-iy

g x^-{-3a^ + 3x*-10a^-x + 6
x* + 3x^

379. If the denominator of a fraction can be resolved
into factors partly of the first and partly of the second
degree, or all of the second degree, in x, and the numerator
is of a lower degree than the denominator, the Theorem of
Undetermined Coefficients enables lis to express the given
fraction as the sum of two or more partial fractions, whose
denominators are factors of the given denominator, and
whose numerators are independent of x in the case of
fractions corresponding to factors of the first degree, and
of the form Ax + B in the case of fractions corresponding
to factors of the second degree.

1. Separate into partial fractions.

or + 1

The factors of the denominator are x + 1 and x'^ — x + 1.

Assume then -^— = ^- + ^^^ + ^ • (1)

X^+l X+ I X2 — X+1

Clearing of fractions, 1 = A(x'^ - x + 1) + (Bx + C)(x+ 1).
Or, 1 = (^ + -B)x2 + (-A + B+C)x + A+C.

Equating the coefficients of like powers of x,

A + B = 0.
-A + B+C^O.

A+ 0=1.

330

ALGEBRA.

Solving these equations, we have
1

A = -, B
3

Substituting in (1),

-, and C = ?-
3 3

1 X

a;3 + l 3(x+l) 3(x2-x+l)

, Am

EXAMPLES.
Separate each of the following into partial fractions

2.

cc^ + l '

a;2 -I- 16 re - 12
(3 a; + 1) (a;' -a; + 3)
2 0!^ + 11 a; - 7

6.

12 +13 x-2a?

8 a;« - 27

2 3^ + 2x^^ + 10

x^ + x^ + 1

REVERSION OF SERIES.

380. To revert a given series ?/ = a + ftx"* + ex" -| is to

express x in the form of a series proceeding in ascending
powers of y.

1. Eevert the series ?/ = 2x + x^ — 2x^ — 3x*+ •••.

Assume x = Ay + By" + Cy^+ Dy* + •••. (1)

Substituting in this the given value of ?/, and performing the opera-
tions indicated, we have

x = A(2x + x"^ - 2x^ - Sx* + •■•)

+ 5(4a;2 + a;* + 4a;3-8a;*+ ••.)
+ C(8x3 + 12x4 + -..)+Z>(16x*+ •••)+—.
X* + •••.

That is, X = 2 ^x + A

x'^-2A

x^- SA

+ iB

+ 4B

- IB

+ 8C

+ 12(7
+ 16Z>

Equating the coeflScients of like powers of x,

2^ = 1.

A + 4B = 0.

-2A + 4B + 8C = 0.

^3A-1B + U

C+WD

= 0 ; etc.

UNDETERMINED COEFFICIENTS. 331

Solving these equations,

A = -, B = --, C = ~, D = -—, etc.
2 8 16 128

113 13

Substituting in (1), ^ ^ 2^ - ^V'^ + ^6^^ ~ 128^* '^ ""' ■^"*'

If the even powers of x are wanting in tlie given series,
the operation may be abridged by assuming x equal to a
series containing only the odd powers of y.

EXAMPLES.
Revert each of the following to four terms ;
2. y = X — x"^ -\- x^ — x'^ -\- '••.

/y*^ rytO /y*^

^ 2 3 4

4. y = x + 2x^ + Sx^ + 4.x*-\ .

5. y = x — 3x^-{-5oi:^ — 7x*-\ .

/>»2 /yi3 /yA

6. y = x--+^--+':.

[2 li li

/y* -T*" /T**^ ^^

^ 2 4 6 8
8. y = x + s(^ + 2x^ + 5x''^

/y»3 /^5 /yti

^ 3 5 7

332 ALGEBRA.

XXXIII. THE BINOMIAL THEOREM.

FRACTIONAL AND NEGATIVE EXPONENTS.

381. It was proved in § 359 that, if n is a positive integer,

(1 4- a:)" = 1 + nx + <^\-^)x^ + n(n-l)(n-^^ _^ ,,^_ ^^^

|w [3

382. Proof of the Theorem for a Fractional or Negative
Exponent.

I. When the exponent is a positive fraction.

p ...

Let the exponent be -, where p and q are positive integers.

By § 211, (1 + x)^ = V(l + xy = i/l+px+..., by (1).

It is evident that a process may be found, analogous to
those of §§ 194 and 200, for expanding -^l+px+--' in
ascending powers of x ; and the first term of the result will
evidently be 1.

Assume then, VT+paT+TTT = 1 + J/x + Nof -\ . (2)

Eaising both members to the qth. power, we have

1 +px + ... = [1 + (Mx + JSfx^ + •••)]'

= 1 +q(3fx + Nx' + •••)+ •-, by (1).

This equation is satisfied by every value of x which makes
both members convergent; and by the Theorem of Unde-
termined Coefficients (§ 369), the coefficients of x in the two
series are equal.

That is, p = qM, or M=^'

Substituting this value in (2), we have

{l + xy=l-\-lx+.... (3)

THE BINOMIAL THEOREM. 333

II. When the exponent is a negative integer or a negative
fraction.

Let the exponent be — s, where s is a positive integer or
a positive fraction.

By § 214, (1 + .)- = ^= ^^-, by (1) or (3).

It is evident that can be expanded by division

l-\-sx-{ -^

in a series proceeding in ascending powers of x ; thus,

1 + SX+ --OlCl —SX+ ...
1 + sx + ...

— sx ~ . . •
That is, (1 + x)-' = 1- sx-\- .... (4)

From (3) and (4), we observe that, when n is fractional
or negative, the form of the expansion is

{1 -\- xy = 1 -\- nx + Ax^ + Boer -] . (5)

X

Writing - in place of a*, we obtain

aj a a- a'^

Multiplying both members by a",

(a + xy = a" + «a"-'x' + Aa^'-'^ci? + Ba^'-^a? + ••-. (6)

This result is in accordance with the second, third, and
fourth laws of § 357 ; hence, these three laws hold for frac-
tional or negative values of the exponent.

We will now prove that the fifth law of § 357 holds for
fractional or negative values of the exponent.

Let P denote the coefficient of x^, and Q the coefficient of
a;'"+\ in the second member of (5).

Then (5) and (6) may be written

(1 -\- xy = 1 -{- nx + •■■ + Px^ + Qx'+i + ..., (7)

and (a + a;)" = a" + na^^^x + . . .

+ Pa"-'a;' + Qa"-' ^x'+^ + • • .. (8)

334 ALGEBRA.

In (8) put a = 1 + 2/ and x = z\ then,

(1 + 2/ + 2=)" = (1 + 2/)" + - + ■?(! + yy^' + •••• (9)
Again, in (7) put x = z-\-y\ then,

(1 + 2 + 2/)" = 1 + ••• + P(2 + 2/)" + Q (^ + 2/r' + -.

Expanding the powers of 2; + ?/ by aid of (8), we have

(1 + 2; + 2/)» = 1 + ••• + PC^-- + rz^-^y + ...]

+ Q[2'-+i + (r + lK2/ +•••] + -. (10)

Since the first members of (9) and (10) are identical, their
second members must be equal for every value of z which
makes both series convergent ; and by the theorem of Unde-
termined Coefficients, the coefficients of z"" in the two series
are equal.

Or, P(l + ?/)"-'• -: P + Q (r + l)y + terms in y\ y\ etc.

Expanding the first member by aid of (7), this becomes

P[l + (n _ r)y + •••] == P+ Q(r + l)y +-.

This equation is satisfied by every value of y which makes
both members convergent, and hence the coefficients of y in
the two series are equal.

That is, P{n - r) = Q(r + 1), or Q = ^i"^ -'>').

But in the second member of (8), n — r is the exponent
of a in the term whose coefficient is P, and r + 1 is the
exponent of x in that term increased by 1.

Hence, the fifth law of § 357 has been proved to hold for
fractional or negative values of the exponent.

By aid of the fifth law, the coefficients of the successive
terms after the second, in the second member of (8), may
be readily found as in § 358 ; thus,

(a + a;)" = a" + na'^-'x + ^<^^~^)a"-V

_^«(«^-J0(nzi2)an-3^+.... (11)

THE BINOMIAL THEOREM. 335

The second member of (11) is an in'&nite series ; for if n
is fractional or negative, no one of the quantities n — 1,
w — 2, etc., can become equal to zero.

The result expresses the value of (a + x)" only for such
values of a and x as make the series convergent (§ 367).

EXAMPLES.

383. In expanding expressions by the Binomial Theorem
when the exponent is fractional or negative, the exponents
and coefficients of the terms may be obtained by aid of the
laws of § 357, which have been proved to hold universally.

If the second term of the binomial is negative, it should
be enclosed, sign and all, in a parenthesjs before applying
the laws ; if either term has a coefficient or exponent other
than unity, it should be enclosed in a parenthesis before
applying the laws.

1. Expand (a + xy to four terms.

The exponent of a in the first term is | ; in the second term, — i ; in
the third term, — | ; in the fourth term, — | ; etc.

The exponent of x in the second term is 1 ; in the third term, 2 ;
in the fourth term, 3 ; etc.

The coefficient of the first term is 1 ; of the second term, f ; multi-
plying the coefficient of the second term, f, by — |, the exponent of
a in that term, and dividing the result by the exponent of x in the
term increased by 1, or 2, we have — \ as the coefficient of the third
term ; and so on.

Then, (a + x) » = a^ + | a^z - \ a'^x'^ + j\ a'^x^ , Ans.

2. Expand (1 — 2x~^)-^ to five terms.

(1 - 2x~2)-2 = [1 +(_ 2x"^)]-2

= 1-2 _ 2. 1-3. (-2x"^)+3.1-<. (-2x~h^

-4.1-5. (_2x"i)3 + 5.1-«. (-2x"2)*- '-'
= 1 -f 4 x~^ + 12 x-i + 32 x~2 + 80 x-2 + ..., Ans.

336 ALGEBRA.
3. Expand to five terms.

- if («-^)"'^ (3x^3 + _3^5_ (a-i)-V(3xb* - ■-

1 4 1 7 2 10 13. 4

= o3 _ o!3a;3 + 2a3x3 - -i/a 3 « + 3_5 ^jj s ^3 + ..., ^ns.
Expand each of the following to five terms :
4. {a + x)i. 10. {xi-2y)\ 15.

5. {l-\-x) \ f rrh-*

11. (m-2+V ) • . .N-3

6. (l-a;)-3. V 3/16 (^~h-\ ^

, , , , ' \b a)

7. </a-x. 12. (a*-2x-^p.

8. -J—. ,,1 17. ^[(.^'"^-32/^)-].

9. -A_. , 18. f-^-^^y.

(a-ft/ 14. (x-^ + 2yzy. Wa" J

384. The formula for the ?*th term of (a + x)" (§ 361)
holds for all values of n, since it was derived from an
expansion which has been proved to hold universally.

EXAMPLES.
1. Find the 7th term of (a — 3x~-^)~k

We have, (a - 3«~')~^ =[« + (- 3x~2)]~^.

In this case, n = — I and j- = 7. _

_3

The exponent of (— Sx -) is 7 — 1, or 6.
The exponent of a is — i — 6, or — J/.

The first factor of the numerator is — i, and the last factor
_ 1^ + 1, or - -V.

The last factor of the denominator is 6.

THE BINOMIAL THEOREM. 337

Hence, the 7th term

_1 4 _7 10 13 16

3 * 3 ' 3 ■ 3 ■ 3 ■ 3 -19 _3
a 3 ( — 3 a; if )*

~ 1.2.3.4.5.6

70ft 1 8 ■ 1 9

Note. The note to § 362 applies with equal force to the examples
in the present article.

Find the

2. 5th term of (a + x)\ 7. 5th term of

3. 7th term of (a + &)~i

V(l - xf

4. 12th term of (1 - x)-\ 8. 7th term of («•* - x^f.

5. ethtermof (a;->+22/*)-2.

r 9. 10th term of

6. 9th term of (a + 2 xy. {x + mf

10. 8th term of (m*- 271-")"^.

11. 9th term of V(a — xy.

12. 6th term of (a' - h-"")'^.

13. 8th term of (;«-3 + 3 y~^)~^.

14. 10th term of (x Vf - -y^X*'

15. 11th term of (a^ + 3 b~^)i.

385. Extraction of Roots by the Binomial Theorem.

1. Find V25 approximately to five places of decimals.
We have, y/25 = 25^ = (27 - 2)^ = (3^ - 2)i

Expanding by the Binomial Theorem, we have

[(33) + (-2)]*=(33)U^(3T'(-2)-l(.S3)'t(_2)3

338 ALGEBRA.

Or, </m = 3--^ + ^ ^0

3.32 9.35 81.38

Expressing the value of each fraction approximately to the nearest
fifth decimal place, we have

\/25 = 3 - .07407 - .00183 - .00008 = 2.92402 ... , Ans.

Rule.

Separate the given number' into two parts, the first of which
is the nearest perfect power of the same degree as the required
root.

Expand the result by the Binomial Tlieorem.

Note. If the second term of the binomial is small compared with
the firet, the terms of the expansion diminish rapidly ; but if the
second term is large compared with the first, it rsquires a great many
terms to ensure any degree of accuracy.

EXAMPLES.

Find the approximate value of each of the following
to five places of decimals :

2. V26. 4. ^. 6. </l7.

3. V98. 6. -^78. 7. -\/^.

LOGARITHMS. 339

XXXIV. LOGARITHMS.

386. Every positive number may be expressed, exactly
or approximately, as a power of 10.

Thus, 100 = 102; i3 = ioi"39-; etc.

When thus expressed, the corresponding exponent is
called its Logarithm to the Base 10.

Thus, 2 is the logarithm of 100 to the base 10 ; a relation
which is written logio 100 = 2, or simply log 100 = 2,

387. Logarithms of numbers to the base 10 are called
Common Logarithms, and, collectively, form the Common
System.

They are the only ones used for numerical computations.

Any positive number, except unity, may be taken as the
base of a system of logarithms ; thus, if a^ = m, where
a and m are positive numbers, then x = log^ m.

Note. A negative number is not considered as having a logarithm.

388. By §§ 213 and 214,

10" = !, 10-^ =!-=.!,

10^ = 10, io-2 = A_=.oi,

102 = 100, 10-3 = A. = .001, etc.

10

10^
10^

Whence by the definition of § 386,

logl = 0, log.l = -l = 9-10,

log 10 = 1, log.01=-2 = 8-10,

log 100 = 2, log .001 = - 3 = 7 - 10, etc.

Note. The second form for log.l, log .01, etc., is preferable in
practice. If no base is expressed, the base 10 is understood.

340 ALGEBRA.

389. It is evident from § 388 that the logarithm of a
number greater than 1 is positive, and tlie logarithm of a
number between 0 and 1 negative. .

390. If a number is not an exact power of 10, its com-
mon logarithm can only be expressed approximately ; the
integral part of the logarithm is called the characteristic,
and the decimal part the mantissa.

For example, log 13 = 1.1139.

In this case, the characteristic is 1, and the mantissa
.1139.

For reasons which will appear hereafter, only the man-
tissa of the logarithm is given in a table of logarithms of
numbers ; the characteristic must be found by aid of the
rules of §§ 391 and 392.

391. It is evident from § 388 that the logarithm of a
number between

1 and 10 is equal to 0 -f a decimal ;
10 and 100 is equal to 1 + a decimal ;
100 and 1000 is equal to 2 + a decimal ; etc.

Therefore, the characteristic of the logarithm of a num-
ber with one figure to the left of the decimal point is 0 ;
with two figures to the left of the decimal point is 1 ; with
three figures to the left of the decimal point is 2 ; etc.

Hence, the characteristic of the logarithm of a numher
greater than 1 is 1 less than the number of places to the left
of the decimal point.

For example, the characteristic of log 906328.51 is 5.

392. In like manner, the logarithm of a number between

1 and .1 is equal to 9 + a decimal — 10
.1 and .01 is equal to 8 + a decimal — 10
.01 and .001 is equal to 7 + a decimal — 10 ; etc.

LOGARITHMS. 341

Therefore, the characteristic of the logarithm of a deci-
mal with no ciphers between its decimal point and first
significant figure is 9, with — 10 after the mantissa ; of a
decimal with one cipher between its point and first signifi-
cant figure is 8, with — 10 after the mantissa; of a decimal
with tivo ciphers between its point and first significant fig-
ure is 7, with — 10 after the mantissa ; etc.

Hence, to find the characteristic of the logarithm of a num-
ber less than 1, subtract the number of ciphers between the
decimal point and first significant figure from 9, writing — 10
after the mantissa.

For example, the characteristic of log .007023 is 7, with
— 10 written after the mantissa.

Note. Some wi-iters combine the two portions of the charac-
teristic, and write the result as a negative characteristic before the
mantissa.

Thus, instead of 7.6036 — 10, the student will frequently find
3.6036, a minus sign being written over the characteristic to denote
that it alone is negative, the mantissa being always positive.

PROPERTIES OF LOGARITHMS.

393. In any system, the logarithm of 1 is 0.

For by § 213, a^ = 1 ; whence by § 387, log.l = 0.

394. In any system, the logarithm of the base is 1.
For, a} = a; whence, log„ a = 1.

395. In any system tvhose base is greater than 1, the loga-
rithm ofOis—c/D.

For if a is greater than 1, a-"= — = - = 0 (§ 295).

Whence by § 387, log, 0 = -^.

Note. No literal meaning can be attached to such a result as
loga 0 = — CO ; it must be interpi-eted as follows :

If, in any system whose base Is greater than unity, a number ap-
proaches the limit 0, its logarithm is negative, and increases without
limit in absolute value. (Compare Note to § 296.)

342 ALGEBRA.

396. In any system, the logarithm of a product is equal to
the sum of the logarithms of its factors.

Assume the equations

a' = m] , , _ oorr f ^ = log. m,

; whence by § 387, , ^^ '

a" = W J "^ [y = lOga 71.

Multiplying the assumed equations,

a" X a" = mn, or a*+^ = mn.

Whence, log„ mn z= x + y = log„ vi + logo n.

In like manner, the theorem may be proved for the prod-
uct of three or more factors.

397. By aid of § 396, the logarithm of a composite num-
ber may be found when the logarithms of its factors are
known.

1. Given log 2 = .3010 and log 3 = .4771; find log 72.

log 72 = log (2 X 2 X 2 X 3 X 3)

= log 2-1- log 2 + log 2 -I- log 3 + log 3 (§396)

=s 3 X log2 -I- 2 X logs = .9030 + .9542 = 1.8572, Ans.

EXAMPLES.

Given log 2 = .3010, log 3 = .4771, log 5 = .6990, and
log 7 = .8451, find:

2. log 35. 7. log 126. 12. log 324. 17. log 1125.

3. log 50. 8. log 196. 13. log 378. 18. log 2625.

4. log 42. 9. log 245. 14. log 405. 19. log 6048.

5. log 75. 10. log 210. 15. log 875. 20. log 12005.

6. log 40. 11. log 625. 16. log 686. 21. log 15876.

398. In any system, the logarithm of a fraction is equal to
the logarithm of the numerator minus the logarithm of the
denominator.

LOGARITHMS. 343

Assume the equations

; wlience, \ , ' '

a'' = 171

a" = n
Dividing the assumed equations,

a" m m

— = —, or a' " = —
a" n n

7th

Whence, log,, — = x — y = log„ m — log^ n.

399. 1. Given log 2 = .3010 ; find log 5.

log 5 = log V- = log 10 - log 2 (§ 398) = 1 - .3010 = .6990, Ans.

EXAMPLES.
Given log 2 = .3010, log 3 = .4771, and log 7 = .8451, find :

2. logJ^. 5. log 45. 8. log If. 11. log28|-.

3. log|. 6. log If 9. log6f 12. log-^^fo.

4. logl4f 7. log 225. 10. log 135. 13. logllO^

400. In any system, the logarithm of any power of a quan-
tity is equal to the logarithm of the quantity multijjlied by the
exponent of the power.

Assume the equation a' = m; whence, x = log^ m.

Raising both members of the assumed equation to the
pth. power,

a*" = m^ ; whence, log„ m^ =px=p log^ m.

401. In any system, the logarithm of any root of a quantity
is equal to the logarithm of the quantity divided by the index
of the root.

rr~ - 1

For, log,, V m = log„ (m') = - log„ m (§ 400).

402. 1. Given log 2 = .3010 ; find log 2l

log2i =1 X log2 (§ 400) = I X. 3010 = .5017, Ans.

O O

344 ALGEBRA.

Note. To multiply a logarithm by a fraction, multiply first by the
numerator, and divide the result by the denominator.

2. Given log 3 = .4771 ; find log ^3.

log</3J-^ = --^J^ = mm, Ans.

o o

EXAMPLES.
Given log 2 = .3010, log 3 = .4771, and log 7 = .8451, find :

3. log3^ 7. log35^ 11. log24l 15. log a/ 105.

4. log5«. 8. log28i 12. log a/3. 16. log -^^.

5. log2l 9. log27i 13. log-s/5. 17. log -x/98.

6. log7l 10. logisl 14. log ^7. 18. log \/i08.
19. Find log (2^ x 3'^).

By § 396, log (23 x ot ) = log 2^ + log sl = J- log 2 + | log 3
= .1003 + .5964 = .6967, Ans.

Find tlie values of the following:

20. log^^- 22. log (2^x105). 24. log ^- 26.log

21. log/'^y. 23. log 7^. 25. log^^. 27. log

7^

403. To prove the relation

- log. m
logj m = -'

Assume the equations

a' = 171] ( x = loga w,

\ ; whence, \
b^ = m } [y = logj m.

[ ; whence.
From the assumed equations, a' = b".

X

Taking the yth. root of both members, a^ = 6.
Therefore, log„ 6 = -, or 3/ =

y

That is, logj m =

y log„ b

log<.m

10ga&

LOGARITHMS. 345

404. To prove the relation

logj a X log„ 6 = 1.
Putting m = a in the result of § 403, we have

-- = 111 = 10^ «^«*>
Whence, logj, a x log^ 6 = 1.

405. In the common system, the mantissce of the logarithms
of numbers having the same sequence of figures are equal.

Suppose, for example, that log 3.053 = .4847.

Then, log 305.3 = log (100 x 3.053)= log 100 + log 3.053

= 2 + .4847 = 2.4847 ;
log .03053 = log (.01 x 3.053) = log .01 + log 3.053

= 8 - 10 + .4847 = 8.4847 - 10 ; etc.

It is evident from the above that, if a number be multi-
plied or divided by any integral power of 10, producing
another number with the same sequence of figures, the
mantissse of their logarithms will be equal.

The reason will now be seen for the statement made in
§ 390, that only the mantissse are given in a table of logar
rithms of numbers.

For, to find the logarithm of any number, we have only
to take from the table the mantissa corresponding to its
sequence of figures, and the characteristic may then be pre-
fixed in accordance with the rules of §§ 39i and 392.

Thus, if log 3.053 = .4847, then

log 30.53 = 1.4847, log .3053 =9.4847-10,

log 305.3 = 2.4847, log .03053 =8.4847-10,

log 3053. = 3.4847, log .003053 = 7.4847 - 10, etc.

This property is only enjoyed by the common system of
logarithms, and constitutes its superiority over others for
the purposes of numerical compatation.

346 ALGEBRA.

406. 1. Given log 2 = .3010, log 3 = .4771 ; find log .00432.
We have, log 432 = log (2* x S^) = 4 log 2 + 3 log 3 = 2.6353.
Then by § 405, the ^nantissa of the result is .6353.

Whence by § 392, log .00432 = 7.6353 - 10, Atis.

EXAMPLES.
Given log 2 = .3010, log 3 = .4771, and log 7 = .8451, find :

2. log 2.8. 7. log .00375. 12. log 2.592.

3. log 11.2. 8. log 6750. 13. log 274.4.

4. log .63. 9. log .0392. 14. log (3.5)«.
■ 5. log .098. 10. log .000343. 15. log a/61.

6. log 32.4. 11. log .875. 16. log (12.6)1

USE OF THE TABLE.

407. The table (pages 348 and 349) gives the mantissae
of the logarithms of all integers from 100 to 1000, calculated
to four places of decimals.

408. To find the logarithm of a number of three figures.

Look in the column headed "No." for the first two sig-
nificant figures of the given number.

Then the mantissa required will be found in the corre-
sponding horizontal line, in the vertical column headed by
the third figure of the number.

Finally, prefix the characteristic in accordance with the
rules of §§ 391 or 392.

For example, log 168 = 2.2253 ;

log .344 = 9.5366 - 10 ; etc.

409. For a number consisting of one or two significant
figures, the column headed 0 may be used.

Thus, let it be required to find log 83 and log 9.

LOGARITHMS. 347

By § 405, log 83 has the same mantissa as log 830, and
log 9 the same mantissa as log 900.

Hence, log 83 = 1.9191, and log 9 = 0.9542.

410. To find the logarithm of a number of more than three

fil/ares.

Let it be required to find the logarithm of 327.6.

From the table, log 327 = 2.5145,
and log328 = 2.5159.

That is, an increase of one unit in the number produces
an increase of .0014 in the logarithm.

Therefore, an increase of .6 of a unit in the number will
produce an increase of .6 x .0014 in the logarithm, or .0008
to the nearest fourth decimal j)lace.

Whence, log 327.6 = 2.5145 + .0008 = 2.5153.

Note. The difference between any mantissa in the table and the
mantissa of the next higher number of three figures is called the tab-
ular difference. The subtraction may be performed mentally.

The following rule is derived from the above :

Find from the table the mantissa of the first three significant
figures, and the tabular difference.

Multiply the latter by the remaining figures of the number,
with a decimal point before them.

Add the result to the mantissa of the first three figures, and
prefix the proper characteristic.

EXAMPLES.
4U. 1. Find log .021508.

Tabular difference = 21 Mantissa of 215 = 3324

.08 2

Correction = 1.68 = 2, nearly. 3326

^ Result, 8.3326 - 10,

348

ALGEBRA.

No.

0

1

2

3

4

6

6

7

8

9

lO

0000

0043

0086

0128

0170

0212

0253

0294

0334

0374

II

0414

0453

0492

0531

0569

0607

0645

0682

0719

0755

12

0792

0828

0S64

0899

0934

0969

1004

1038

1072

1106

13

"39

"73

1206

1239

1271

1303

1335

1367

1399

1430

14

1461

1492

1523

1553

1584

1614

1644

1673

1703

1732

15

1761

1790

1818

1847

1875

1903

1931

1959

1987

2014

16

2041

2068

2095

2122

2148

2175

2201

2227

2253

2279

17

2304

2330

2355

2380

2405

2430

2455

2480

2504

2529

18

2553

2577

2601

2625

2648

2672

2695

2718

2742

2765

19

2788

2810

2833

2856

2878

2900

2923

2945

2967

2989

20

3010

3032

3054

3075

3096

3118

3139

3160

3181

3201

21

3222

3243

3263

3284

3304

3324

3345

3365

3385

3404

22

3424

3444

3464

3483

3502

3522

3541

3560

3579

3598

23

3617

3636

3655

3674

3692

37"

3729

3747

3766

3784

24

3802

3820

3838

3856

3874

3892

3909

3927

3945

3962

25

3979

3997

4014

4031

4048

4065

4082

4099

4116

4133

26

4150

4166

4183

4200

4216

4232

4249

4265

4281

4298

27

4314

4330

4346

4362

4378

4393

4409

4425

4440

4456

28

4472

4487

4502

4518

4533

4548

4564

4579

4594

4609

29

4624

4639

4654

4669

46S3

4698

4713

4728

4742

4757

30

4771

4786

4S00

4814

4829

4843

4857

4871

4886

4900

31

4914

4928

4942

4955

4969

4983

4997

501 1

5024

5038

32

5051

5065

5079

5092

5105

5"9

5132

5145

5159

5172

33

5185

5198

5211

5224

5237

5250

5263

5276

5289

5302

34

5315

5328

5340

5353

5366

5378

5391

5403

5416

.5428

35 5441

5453

5465

5478

5490

5502

5514

5527

5539

5551

36

5563

5575

5587

5599

561 1

5623

5635

5647

5658

5670

37

5682

5694

5705

5717

5729

5740

5752

5763

5775

5786

38

5798

5809

5821

5832

5843

5855

5866

5877

5888

5899

39

59"

5922

5933

5944

5955

5966

5977

5988

5999

6010

40

6021

6031

6042

6053

6064

6075

6085

6096

6107

6'v'

41

6128

6138

6149

6160

6170

6180

6191

6201

6212

42

6232

6243

6253

6263

6274

6284

6294

6304

6314

63^.

43

6335

6345

6355

6365

6375

6385

6395

6405

6415

6425

44

6435

6444

6454

6464

6474

6484

6493

6503

6513

6522

45

6532

6542

6551

6561

6571

6580

6590

6599

6609

6618

46

6628

6637

6646

6656

6665

6675

6684

6693

6702

6712

47 1 6721

6730

6739

6749

6758

6767

6776

6785

6794

6803

48 6812

6821

6830

6839

6S48

6857

6866

6875

68S4

6S93

49

6902

691 1

6920

6928

6937

6946

6955

6964

6972

6981

50

6990

6998

7007

7016

7024

7033

7042

7050

7059

7067

51

7076

7084

7093

7101

7110

7118

7126

7135

7H3

7152

52

7160

7168

7177

7185

7193

7202

7210

7218

7226

7235

53

7243

7251

7259

7267

7275

7284

7292

7300

7308

7316

54

7324

7332

7340

7348

7356

7364

7372

7380

7388

7396

No.

0

1

2 i 3

4

5

6

7

8

9

LOGARITHMS.

849

No.

0

1

2

3

4

5

6

7

8

0

55

7404

7412

7419

7427

7435

7443

7451

7459

7466

7474

56

7482

7490

7497

7505

7513

7520

7528

7536

7543

7551

57

7559

7566

7574

7582

7589

7597

7604

7612

7619

7627

58

7634

7642

7649

7657

7664

7672

7679

7686

7694

7701

59

7709

7716

7723

7731

7738

7745

7752

7760

7767

7774

60

7782

7789

7796

7803

7810

7818

7825

7832

7839

7846

61

7853

7860

786S

7S75

7SS2

7SS9

7S96

7903

7910

7917

62

7924

7931

7938

7945

7952

7959

7966

7973

7980

7987

63

7993

8000

8007

8014

8021

8028

8035

8041

8048

8055

64

8062

8069

8075

8082

8089

8096

8102

8109

8116

8122

65

8129

8136

8142

8149

8156

8162

8169

8176

8182

8189

66

8195

8202

8209

8215

8222

822S

8235

8241

8248

8254

67

8261

8267

•8274

8280

8287

8293

8299

8306

8312

8319

68

8325

8331

8338

8344

8351

8357

8363

8370

8376

8382

69

8388

8395

8401

8407

8414

8420

8426

8432

8439

8445

70

8451

8457

8463

8470

8476

8482

8488

8494

8500

8506

71

8513

8519

8525

8531

8537

8543

8549

8555

8561

8567

72

8573

8579

8585

8591

8597

8603

8609

8615

8621

8627

73

8633

8639

8645

8651

8657

8663

8669

8675

8681

8686

74

8692

8698

8704

8710

8716

8722

8727

8733

8739

8745

75

8751

8756

8762

8768

8774

8779

8785

8791

8797

8802

76

8808

8814

8820

8825

8831

8837

8842

8848

8854

8859

77

8865

8871

8876

8882

8887

8893

8899

8904

8910

8915

78

8921

8927

8932

8938

8943

8949

8954

8960

8965

8971

79

8976

8982

8987

8993

8998

9004

9009

9015

9020

9025

80

9031

9036

9042

9047

9053

9058

9063

9069

9074

9079

81

9085

9090

9096

9101

9106

9112

9117

9122

9128

9133

82

9138

9143

9149

9154

9159

9165

9170

9175

9180

9186

83

9191

9196

9201

9206

9212

9217

9222

9227

9232

9238

84

9243

9248

9253

9258

9263

9269

9274

9279

9284

9289

85

9294

9299

9304

9309

9315

9320

9325

9330

9335

9340

86

9345

9350

9355

9360

9365

9370

9375

9380

9385

9390

87

9395

9400

9405

9410

9415

9420

9425

9430

9435

9440

88

9445

9450

9455

9460

9465

9469

9474

9479

9484

9489

89

9494

9499

9504

9509

9513

9518

9523

9528

9533

9538

90

9542

9547

9552

9557

9562

9566

9571

9576

9581

9586

91

9590

9595

9600

9605

9609

9614

9619

9624

9628

9633

92

9638

9643

9647

9652

9657

9661

9666

9671

9675

9680

93

9685

96S9

9694

9699

9703

9708

9713

9717

9722

9727

94

9731

9736

9741

9745

9750

9754

9759

9763

9768

9773

95

9777

9782

9786

9791

9795

9800

9805

9809

9814

9818

96

9823

9827

9S32

9836

9841

9845

9850

9854

9859

9863

97

986S

9872

9877

9881

9886

9890

9894

9899

9903

9908

98

9912

9917

9921

9926

9930

9934

9939

9943

9948

9952

99

9956

9961

9965

9969

9974

9978

9983

9987

9991

9996

No.

0

1

2

3

4

5

6

~7~

8

9

350 ALGEBRA.

Find the logarithms of the following:

2. 53. 6. 1068. 10. 7.803. 14. 4072.6.

3. 2.6. 7. 82.95. 11. .0003787. 15. .0064685.

4. 871. 8. .9616. 12. 253.07. 16. .013592.

5. .689. 9. .007254. 13. .91873. 17. 4.0354.

412. To find the number corresponding to a logarithm.

1. Required the number whose logarithm is 1.6571.

Find in the table the mantissa 6571.

In the corresponding line, in the column headed "No.,"
we find 45, the first two figures of the required number, and
at the head of the column we find 4, the third figure.

Since the characteristic is 1, there must be two places to
the left of the decimal point (§ 391).

Hence, the number corresponding to 1.6571 is 45.4.

2 Required the number whose logarithm is 2.3934.

We find in the table the mantissas 3927 and 3945, whose
corresponding numbers are 247 and 248, respectively.

That is, an increase of 18 in the mantissa produces an
increase of one unit in the number corresponding.

Therefore, an increase of 7 in the mantissa will produce
an increase of -J-g of a unit in the number, or .39, nearly.

Hence, the number corresponding is 247 + -39, or 247.39.

The following rule is derived from the above :

Find from the table the next less mantissa, the three figures
corresponding, and the tabidar difference.

Subtract the next less from the given mantissa, and divide
the remainder by the tabular difference.

Annex the quotient to the first three figures of the number,
and point off the result.

Note. The rules for pointing off are the reverse of the rules for
characteristic given in §§ 391 and 392.

LOGARITHMS. 351

I. If — 10 is not written after the mantissa, add 1 to the character-
istic, giving the number of places to the left of the decimal point.

II. If — 10 is written after the mantissa, subtract the positive part
of the characteristic from 9, giving the number of ciphers to be placed
beticeen the decimal point and first significant figure.

EXAMPLES.

413. 1. Pind the number whose logarithm is 8.5264 — 10.

5264
Next less mantissa = 5263 ; three figures corresponding, .336.

Tabular difference, 13)1.000(.077 = .08, nearly.
91_
90
According to the rule of § 412, there will be one cipher between the
decimal point and first significant figure.

Hence, the number corresponding = .033608, Ans.

Find the numbers corresponding to the following loga-
rithms :

2.

0.3075.

7.

9.9108 -

-10.

12.

7.5862 -

-10.

3.

8.7284 -

-10.

8.

7.6899 -

-10.

13.

9.7043 -

-10.

4.

1.8079.

9.

0.8744.

14.

2.5524.

5.

3.3565.

10.

8.9645 -

-10.

15.

4.2306.

6.

2.6639.

11.

1.8077.

16.

6.2998 -

-10.

APPLICATIONS.

414. The approximate value of an arithmetical quantity,
in which the operations indicated involve only multiplica-
tion, division, involution, or evolution, may be conveniently
found by logarithms.

The utility of the process consists in the fact that addi-
tion takes the place of multiplication, subtraction of division,
multiplication of involution, and division of evolution.

Note. In computations with four-place logarithms, the results
cannot usually be depended upon to more than fo^lr significant fig-
ures,

352 ALGEBRA.

415. 1. Find tlie value of .0631 x 7.208 x .51272.

By § 396,

log (.0631 X 7.208 x .51272) = log .0631 + log 7.208 + log. 61272.

log .0631 = 8.8000 - 10

log 7.208= 0.8578

log .51272= 9.7099-10

Adding, log of result = 19.3677 - 20 = 9.3677 - 10. (See Note 1.)

Number corresponding to 9.3677 - 10 = .2332, Ans.

Note 1. If the sum is a negative logarithm, it should be written
in such a form that the negative portion of the characteristic may
be - 10.

Thus, 19.3677 - 20 is written in the form 9.3677 - 10.

2. Find the value of — — V-

7984

By § 398, log ^ = log 336.8 - log 7984.

log336.8 = 12.5273 -10 (See Note 2.)

log 7984 = 3.9022

Subtracting, log of result = 8.6251 - 10

Number corresponding = .04218, Ans.

Note 2. To subtract a greater logarithm from a less, or to sub-
tract a negative logarithm from a positive, increase the characteristic
of the minuend by 10, writing — 10 after the mantissa to compensate.

Thus, to subtract 3.9022 from 2.5273, write the minuend in the form
12.5273 - 10 ; subtracting 3.9022 from this, tlie result is 8.6251 - 10.

3. Find the value of (.07396)^

By § 400, log (.07396)6 = 5 x log .07396.

log .07396 = 8.8690 -10
5

44.3450 - 50 = 4.. 3450 - 10 (See Note 1.)

Number corresponding = .000002213, Ans.

LOGARITHMS. 353

4. Find the value of V.035063.

By § 401, log ^.035063 = - log .035063.

log.035063= 8.5449 -10

3)28.5449 - 30 (See Note 3. )

9.6160 - 10
Number corresponding = .3274, Ans.

Note 3. To divide a negative logarithm, write it in such a form
that the negative portion of the characteristic may be exactly divisible
by the divisor, with — 10 as the quotient.

Thus, to divide 8.5449 — 10 by 3, we write the logarithm in the
form 28.5449 - 30. Dividing this by 3, the quotient is 9.5150 - 10.

ARITHMETICAL COMPLEMENT.

416. The Arithmetical Complement of the logarithm of a
number, or, briefly, the Cologarithm of the number, is the
logarithm of the reciprocal of that number.

Thus, colog 409 = log -i- = log 1 - log 409.

log 1 = 10 - 10 (Note 2, § 415.)

log 409= 2.6117

.-. colog 409= 7.3883-10.

Again, colog .067 = log — — = log 1 — log .067.

.067

logl = 10 -10

log .067= 8.8261-10

.-. colog .067= 1.1739.

It follows from the above that the cologarithm of a num-
ber may he found by subtracting its logarithm from 10 — 10.

Note. The cologarithm may be obtained by subtracting the last
significant figure of the logarithm from 10 and each of the others from
9, — 10 being written after the result in the case of a positive loga-
rithm.

854 ALGEBRA.

.61384

417. Example. Find the value of

8.709 X .0946

log -51384 ^ log /.51384 x -1- x ^
8.709 X .0946 V 8.709 .09-

0946

= log .51384 + log — ^ + log ^—
^ 8.709 .0946

= log. 51384 + colog 8.709 + colog.0946.

log.51384 = 9.7109 -10
colog8.709 = 9.0601 -10
colog .0946 = 1.0241

9.7951 - 10 = log .6239, Ans.

It is evident from the above example that the logarithm
of a fraction is equal to the logarithm of the numerator plus
the cologarithni of the denominator.

Or in general, to find the logarithm of a fraction whose
terms are composed of factors,

Add together the logarithms of the factors of the numerator,
md the cologarithms of the factors of the denominator.

Note. The value of the above fraction may be found without
using cologarithms, by the following formula :

log ^IM ^ log. 51384 - log (8.709 x .0946)

^ 8. 709 X. 0946 ° '' ^ ^

= log. 51384 - (log 8.709 + log .0946).

The advantage in the use of cologarithms is that the written work
of computation is exhibited in a more compact form.

EXAMPLES.

Note. A negative quantity has no common logarithm (§ 387,
Note). If such quantities occur in computation, they should be
treated as if they were positive, and the sign of the result determined
irrespective of the logarithmic work.

Thus, in Ex. 3, § 418, the value of 847.5 x (- 2.2807) is obtained
by finding the value of 847.5 x 2.2807, and putting a negative sign
before the result. See also Ex. 34.

LOGARITHMS. 355

418. Find by logarithms the values of the following :

1. 3.142x60.39. 4. (- 4.3918) x (- .070376).

2. 541.21 X .01523. 5. .93653 x .0031785.

3. 847.5 x (- 2.2807). 6. (- .00017435) x 69.571.

7 486.7 g .5394 ^^ 9563.2

76.51' ■ -.09216' ■ 42712*

g 1.0547 ,Q 2.708 ,0 - .00006802

34.946' ■ .0086819' " .0071264

,3 3.8961 X .6945 ,5 (- .87028) x 3.74

■ 4694 X. 00457' ■ (- .06589) x(- 42.318)"

.. 718 x(- .02415) jg .09213 x (- 73.36)

■ (- .5157) X 1420.6' ' .832 x 2808.7 *"

17. (7.795)*. 22. (.7)1 27. ^^^9.

18. (.8328)^ 23. (-964)1 28. ^^/lOO.

19. (-25.14)3. 24. (.00105)1 29. ^3994.

20. (.03512)2. 25. V5. 30. ^^117256.

21. 10*. 26. ■y/2. 31. ■</.002613.

32. -v/-. 00095173.

33. Find the value of ^.
3*

log i^ = log 2 + log v^ + cologS^ (§ 417)

= log 2 + hog 5 + I colog 3.
o o

log 2 = .3010

log 6 = .6990 ; divide by 3 = .2330

oolog 3 = 9.5229 - 10 ; multiply by - = 9.6024 - 10

6 .

.1364 = log 1.369, Ans,

356 ALGEBRA.

34. Find the value of

s/- .03296
\ 7.962

log ;pi96 ^ 1 1 ,03296 ^ 1 ^ ^3296 - log 7.962).
^^ 7.962 3 ^ 7.962 3^ ° & v

log .03296 = 8.5180 - 10
log 7.962 = 0.9010

3)27.6170 - 30

9. 2057 -10 = log. 1606.

Result, - .1606.

Find the values of the following

35. 4^x7l 38 (-01)^

■</7

39.

36. §!.

37. ^1

45. -v/S X ^ X \/7.

.. ^76 X .0592^^

/M

(-10)^
40. f-MOOy

41. V^^

42.

43.

940
-•v/1000

47

x/^

7.543

31 X .414
48. a/. 0009657

(-.6)^
60. (25.467)1° X (- .062)".

51. ^5106.5 X .00003109.

52. (83.74 X .009433)7.

53. (4.8671)^ X (.17543)i

K- V3J28 X -^^65:47
04.

-v^721.32
(.573)^

55.

49.

-V/.004978
- (.25691)^

8693.8 X ■V/.03307

(- .0001916)^ X V68l
^®- - .27556

(- .83457)^

57. V:374 X -v/:05286 x ^.0078359.
38.014

58.

</:04142 X (- .947')

LOGARITHMS. 357

EXPONENTIAL EQUATIONS.

419. An Exponential Equation is an equation of the form

a" — b.

To solve an equation of this form, take the logarithms of
both members ; the result will be an equation which can be
solved by ordinary algebraic methods.

1. Given 31'' = 23 ; find the value of x.

Taking the logarithms of both members,

log (31^) = log 23.
Whence, a; log 31 = log 23 (§ 400).

Therefore, x = ^^^ = ^^^^ = .91303, Ans.

log31 1.4914 '

2. Given .2"= = 3 ; find the value of x.

Taking the logarithms of both members,
xlog.2 = log 3.

Whence, z=^-^^^ _4771_ ^ .4771 ^ _ ^^^

log. 2 9.3010-10 -.6990

EXAMPLES.

Solve the following equations :

3. 332^ = 5.17. 5. .0158^ = .008295. 7. a^ = 6 V.

3

4. .416^ = 6.72. 6. 5.336' = .744. 8. mV = n.

9. 72^-3 = .02041. 10. .8^'-^ = .4096.

11. Given a, r, and I; derive the formula for n (§ 346).

12. Given a, r, and S ; derive the formula for n.

13. Given a, I, and S ; derive the formula for n.

14. Given r, I, and S ; derive the formula for n.

420. 1. Find the logarithm of .3 to the base 7.

By § 403, log, .3 = l^io^ = Mlll^liO = ^.5229 ^ _ g^gy, Ans.
Iogio7 .8451 .8451

358 ALGEBRA.

EXAMPLES.
Find the values of the following :

2. logalS. 4. log.68 2.9. 6. logi.6.838.

3. log, .9. 5. log.3,.076. 7. log83 5.2.

Examples like the above may be solved by inspection if
the number can be expressed as an exact power of the base.

8. Find the logarithm of 128 to the base 16.

Let logie 128 = X ; then by § 387, 16^ = 128.
That is, (2*)^ - 2", or 2*^ = 2".

Whence by inspection, 4x = 7, or x = -•

Therefore, logie 128 = -, Ans.
4

9. Find the logarithm of 81 to the base 3.

10. Find the logarithm of 32 to the base 8.

11. Find the logarithm of \ to the base 27.

12. Find the logarithm of Jj to the base -^.

APPENDIX.

GRAPHS*

Note. Sections 1-13 may be studied when the class has reached
Article 168 in this book ; the study of sections 14-17 may precede that
of Article 268, and the later sections may be completed when the class
has reached Article 281.

1. A Graph is the representation by means of a line or
curve of some set of facts. It has a wide application, and
is used to show the fluctuations in the price of wheat, the
variations in temperature for a day, the death rate at various
ages, or the essential facts of an algebraical equation.

2. An equation of the first degree in two variables,

We say in algebra that such an equation is indeterminate,
for we can get an indefinite number of values of x and y that
will satisfy it. To get a solution we need only to assign
arbitrarily to x a value and then solve for y. The following
is a set of solutions :

x = 0

y = 3

x = l

y = H

x = 2

y = H

x = 3

y = l

X= 4:

y = o

x = 5

y = "f

The list could be extended

indefinitely.

* By Robert J. Aley, A.M., Ph.D., Professor of Mathematics,
Indiana University. Copyright, 1901, by D. C. Heath & Co.

369

360

ALGEBRA.

3.

Fig.

II

X'-

III

I

-har. + y

-\-x,-y
IV

Y'

Fig. 1.

Axes. Two lines intersecting at right angles, as in
1, are called the Axes of Coordinates. OX, the hori-
zontal one, is the JT-axis, or Axis
of Abscissas; OY, the vertical
one, is the X-axis, or Axis of
Ordinates. 0, the intersection
of the axes, is the origin.

4. Quadrants. The axes divide
the plane into four parts, called
quadrants. These quadrants are
numbered from I to IV, as in
Fig. 1.

5. Coordinates. A point is located when its distance and
direction from each of the axes is known. The distance
from 0 y is the avdistance, or abscissa. The distance from
OX is the y-distance, or ordinate. The two distances con-
stitute the Coordinates of the point. A point is denoted by
the symbol (x, y) where x is the abscissa and y the ordinate.

6. Convention as to Signs. In the representation of points
distances to the right of the F-axis are positive, to the left
negative. Distances above the X-axis are positive, those
below negative. An x in the first or fourth quadrant is +,
in the second or third it is — . A ?/ in the first or second
quadrant is -{-, in the third or fourth it is — . These are
indicated in Fig. 1.

7. Plotting Points. To locate the y
point Pi (3, 4), we measure 3 units to
the right of 0 Y and then measure 4
units up from OX. (See Fig. 2.). The
point P^{—2, 3) is 2 units to the left
of OF and 3 units above OX. The
point P3(— 3, — 1) is 3 units to the
left of or and 1 unit below OX.
The point (+2, — 3) is 2 units to the
right of 0 Y and 3 units below OX.

Pa (-2„3)

P3(-3-l)

0

Pi ( 3,4)
X

P«(2.-3)

Fio. 2.

APPENDIX.

361

Exercises. Plot the following points .

(2,-4), (-1,-5), (-6,3), (4,4), (-3,-5),

(2, - 2) (- 2, 2) (4, - 1) (- 5, - 1) (4, - 3)

(0,5) ' (0,-4)' (3,0) ' (-4,0) ' (0,0) '

8. The Graph of 3 a; + 4 ?/ = 12.

In section 2 we wrote down six solutions to this equation.
In the notation of section o, we can now write these solutions as
the points (0, 3), (1, 2i), (2, li), (3, |),
(4, 0), and (5, — |). These points
may be plotted as in Fig. 3. It will
now be noticed that these points are
in a straight line. The line is called
the graph of the equation. If the
line be produced indefinitely and
the X and y of any point found by
measurement from the graph, the
values thus found will satisfy the
equation. For example, if we select the point Q, we find
that its X, OM, is — 4 and its y, MQ is + 6. These values
satisfy the equation 3 a; + 4 ?/ = 12, for 3(— 4) + 4 (6) = 12.

9. An equation of the first degree in two variables always
has a straight line for its graph.

ax -{- by = c is a. general linear equation in x and y.
A set of solutions is as follows :

Fig. 3.

x = 0,
x = l,

c

y =
y =

Fie. 4.

x = Z, y =

a; = 4, y =
etc..

c — a

b
c — 2a

b
c — 3a

b
c — 4a

b
etc.

b62

ALGEBRA.

We see that a change of 1 in the value of x makes a
change m y. If these points were plotted, they would

appear (Fig. 4) very much like the side view of a uniform
straight stairway, in which the width of the steps is 1, and

the height -. The points are readily seen to be in a straight
line.

A careful plotting of the following equations will do
much to convince the student of the truth of the above.

Exercises. Get five or more solutions to each of the
following, plot the points, and draw the graph of the
equation:

x-\-y = 5,
Ax-\-y = l,

2 X + 3 ?/ = 6,

re + 42/ = 10,

^x-2y = 12.
2 X — y = 5.

10. A shorter way of getting the graph.

Since the linear equation always represents a straight
line, we can draw its graph if we know two points upon it.
In general the two points most easily determined are those
where the graph cuts the coordinate axes. The point on
the X-axis is found by putting y = 0 and solving for x.
The point on the F-axis is found by putting x = 0 and solv-
ing for y.

Example.

If
and if

2x-5y = 10.

y = 0 X = 5,
x = 0 y = -2.

The required graph cuts the X-axis
at (5, 0) and the F-axis at (0, — 2).
Plotting these two points the line is
easily drawn as in Fig. 5.

Fio. 5.

APPENDIX.

363

11. Simultaneous equations.

(1) f 5 a; + 4 2/ = 22 ] (problem 3,

(2) I 3 a; + y= 9 | page 141).

Draw the graphs of these two
equations on the same diagram as
in Fig. 6. It is found that the two
lines intersect at a point P whose
coordinates are (2, 3). The x and y
(2 and 3) of this point is the solu-
tion of the equations.

Fig. 6.

12. Two simultaneous linear equations in x and y have
but one solution. Each equation represents a straight line.
The solution is the point common to both lines ; that is, the
intersection of the lines. Two straight lines can only in-
tersect in one point, so there is but one solution.

Exercises. Solve by drawing graphs and measuring the
coordinates of the point of intersection :
Page 141, Examples 5, 8, 13, 18.
Page 142, Examples 2, 4, 7, 17.
Page 143, Examples 2, 5, 9, 15.

13.

(1) r x+ ^ = 11

(2) Ux + 2y = l]

If we undertake to solve the above equations, we encounter
a difficulty. We find that we can not eliminate x without
also eliminating y at the same time.

If we draw the graphs of these equa-
tions, we find they are represented as
in Fig. 7. The graphs show at once
where the difficulty is. The lines are
parallel and so do not intersect at all.
In the language of mathematics, theyin-
tersect at infinity, which is just another
Fig. t. way of saying they never intersect.

364

ALGEBRA.

Exercises. Draw the graphs of the following sets of
equations :

2x+ y = 4. (2x + 3y = 12

4:X + 2y = 10 \3x-2y = 5

x + 2y = 6 f x = 5

2/ = 3

I x = 5

\2x-y = 7

THE GRAPH OF THE QUADRATIC.

14. In order to get the graph of the quadratic we intro-
duce y, putting the equation in the form ax- -\-bx-{- c = y.
It is readily seen that when y = 0 this equation becomes the
general quadratic in x.

15. Graph ofar'-2a;-3 = 0.

We write or — 2x — 3 = y. Solving this for x in the
usual way, we get

x =

1 ± V4 + 2/.

win

g list of values for y and x are reac

1.

y =

0

x = 3 and — 1

2.

y=-

-1

aj = 2.7 '

' - .7

3.

y = -

-2

iK = 2.4 '

' -4

4.

y = -

-3

x = 2 '

0

5.

y = -

-4

x = l '

1

6.

y =

1

a: = 3.2 '

' -1.2

7.

y =

2

x = 3A '

' -1.4

8.

y =

3

x = 3.6 '

' -1.6

9.

y =

4

a; = 3.8 '

' -1.8

10.

2/ =

5

tc = 4 *

' -2

11.

y =

et

12

a: = 5 *
etc.

' -3

APPENDIX.

365

Plotting these points carefully and connecting them by a
smooth curve, we get the result shown
in Fig. 8. It is seen that the graph in
this case is a curve, and that it cuts the
axis of X in two points. These points
are at distances of 3 and — 1 from the
origin. 3 and — 1 are the two roots of
the quadratic a^— 2a; — 3 = 0.

16. A quadratic always represents a
curve that can be cut in two places by
one straight line.

Draw the graphs of the following equations

(1-4)

Fig. 8.

1. a;^ + 6a; = 7.

2. a;2 + a; = 30.

3. ^o?^lx = 2.

4. 8a;2 + 2a; = 3.

5. :^-lx = m.

6. 6 ar^ - 11 a; = 10.

7. 2a-2 + a; = 6.

8. «2 + 14 a; -I- 48 = 0.

9. ^3?-x = h.

10. ar'^- 3a; -28 = 0.

11. 5a;2-7a; = 0.

17. ar' - 2 a; + 1 = 0.

Write a;2_2a; + l=^.

Solving for x, we have

a; = 1 ± -sly.

y = o

a;=l

y = l

a; = 2 and

0

y = 4.

a; = 3 and

-1

y = 9

a; = 4 and

-2

etc. .

etc.

366

ALGEBRA.

Plotting these points and drawing a smooth curve through
them, we have tlie curve shown in Fig. 9.
This curve does not cross the axis of X,
but touches it at tire point (1, 0). The
iirst member of the given equation being a
perfect square, the equation has two equal
roots. The graph of a quadratic having
equal roots always touches the axis of X
at a distance from 0 equal to one of the
equal roots.
Draw the graphs of

1. a^ — 4a; = — 4.

2. x^ + 6x = -9.

Fio. <).

Exercises.

GRAPHS OF SIMULTANEOUS QUADRATICS.
18. 1. x + y==2.

2. xy = — 15.

In the solution as given on page 251, two auxiliary equa-
tions occur, viz. ^ 3. a; — w ::= 8.

4. x — y = — 8.

In Fig. 10 the various lines of the graph are numbered to
correspond with the numbers of the
equations.

Equations (1) and (2) give a
straight line and the double-
branched curve known as a hyper-
bola. These intersect at the points
P, Q, whose coordinates are (—3,
5) and (5, — 3). These are the
only solutions to the system of
equations. The auxiliary lines
(3) and (4) intersect line (1) in P
and Q, and hyperbola (2) in R and
S. B and S are solutions of the system (3), (4), and (2),
but not of the system (1), (2).

\4

\p/.

/

_jj«

C8-)'
\ / X

/

\^(--2-r^

r \v

/

1 ^

Fir,. 10,

APPENDIX.

'661

19. 1. a;?/ = 12.

2. x'-\-y- = 40.

The auxiliary equations appearing in the solution are ;

3. x-\-y = + 8.

4. x + y — ~8.

5. X — y = -{- 4t.

6. X — y = — 4.

The graphs of all these equations
are shown in Fig. 11 by the corre-
sponding numbers.

The solutions are at the points P,
Q, R, and S.

Exercises. Draw the graphs of
the equations :

Page 251, Examples 3, 4.
Page 253, Examples, 5, 12, 17.
Page 254, Examples, 3, 6.

ANSWERS TO WELLS'S ESSENTIALS OF
ALGEBRA.

§ 5 ; pag-es 4 to 6.

4. 56, 14. 5. A, 49 ; B, 67. 6. 95, 28.

7. A, 64; B, 38. 8. A, f35; B, $58.

9. A,f48;B, f8. 10. 52,33. 11. A, $18; B, $54.

12. Men, 300 ; women, 100 ; children, 25. 13. $0.99.

14. A, $48; B, $32; C, $16. 15. 52,29,87.

16. A, $78; B, $57; C, $95. 17. Watch, $48; chain, $8.

18. 13, 26, 130. 19. $54, $18, $72.

20. Cow, $55; sheep, $18; hog, $11. 21. 36, 19, 72.

22. A, 322 ; B, 186. 23. A, 33 ; B, 50 ; C, 18.

24. A, $96; B, $24; C, $54.

25. Horse, $240; carriage, $192; harness, $24.
26. $25, $5, $125. 27. A, 140; B, 168; C, 204.
28. A, $12; B, $7; C, $19; D, $31. 29. 4, 12, 36, 108.

§ 8; pages 7, 8.

2.

8.

6.

19
30'

10. 0.

15. 64.

19. 0.

23. 40.
21

3.

360.

7.

27
25*

11. 1.
60

16. 324.

20. 8.

24. 1.

4.

46.

8.

5184

. 12. '''.
64

17. 642.

21. 1

5

25. 16.

5.

61
6'

9.

73

so"

13. ^•

50

§ 16;

18. 284.
page 12.

22. 15.
6

26. f .
6

11.

1
12*

12. -

101 J,
35"

§ 19;

t. 3|.
page 14.

14.

-63V

15. 7

12.

12
35*

"• -k

14.

-

24.

15. 20.
9

2 ALGEBRA.

§ 34; pages 20, 21.
2. a - 4 6. 3. - 2 m2 + 3 tt3. 4. - 6 a6 - 11 cd.

6. 9 a - 4 6 - 6 c. 6.4 ?>i-. 1. x-y - z. %. ba'^-iah -2 b'\
9. 3a;3_a;_4. lO. 6a + 46-2c. 11. x» - Sx^y - 2a;?/2 - 3i/3.

12. 0. 13. 3a3-5a2 + 4a-2. 14. 6a2 _ 3ft-2 _ 5^2.

15. aS - «^ 16. 7 x8 + 22 x2 - 14 X - 24.

§ 39 ; page 23.
25. -37 a. 26. 5xy. 27. - 14a2. 28. SinH.

§ 40; pages 24, 25.

2. 4 ^2 _ 3 a _ 20. Z. Sab -6 be + ca. 4.-4 xy.

6. _ 66 + 8c. 6. x3-x2-6x4-7. T. -3x + 3y -3z.

8. 6a-126 + 21c + 2d. 9. - 9a3 + Sa^ - 4a + 3.
10.8x8+11x2-5. 11. 4a-2a2-2a3. 12. 6a2 + 6a6 - 6662.

13. 10x3 _ 6x2 + 9x - 12. 14. Qa^ + 3a^b- 12ab^ -8bK

16. 2x8- 6x2y + x?/2. 16. 7a-&-8c-4d.

17. 5-4x + 7x2-20x8-6x*. 18. x2-4xj/-6j/2 + 7x + 21y.

19. 4 a6 + 10 a* - 11 a8 - 16 a2 - 8 a + 1.

20. x^ - 6 x*y + 6 x82/2 + 11 x2j/8 _ 16 xy^ + y^.

21. 4a2. 22. 2a2_a&. 23.5x8-8x2-9. 24. 7x-6j/.

25.0. 26. 3a + 36 + 3c + 3d. 27. 12a3 - 8a - 7.

§ 43; pages 27, 28.

3. 6a +12 6. 4. 7?tt-3 7i. 5.x + y-3z. G.Sa^-ab.

7. - 2 m2 + n2. 8. 2x-l. 9.a-b + c + d-e. 10. -2a6 + 3.
11. 8x-7. 12. 0. 13. -10. 14. -4. 15. -lOx + 1.
lG.x + y + z. 17. -3n-5. 18.17. 19. 3a-l. 20.0.

21. -2x + y -2z. 22. x-y. 23. 1.

§ 52 ; pages 33 to 35.
3. 6a2 + 29a + 35. 4. 30a2-53a + 8. 5. -32x2- 52x2/- 15 2/2.
6. 28 a262 + 34 06 _ 12. 7. x8 + y^ 8. 10 a^ + 33 a2 - 52 a + 9.

9. 12x8-13x2+19x-12. 10. 5n8+2n2-19n-6. 11. 27a8-868.

12. a2 - 2 a6 + 2 ac + 62-_ 2 6c + c\
18. 12 m5 + 8 m*?i- 31 m8n2- 24 m2n8.

ANSWERS. 3

14. 3 a^+ 5 a:8 - 33 z^'+ 10 a; + 24. 16. m< + m'^n'^ + n*.

16. 16 a* - 1. 17. 63 x^ + 114 x^ + 49 x2 - 16 x - 20.

18. 8 n* - 50 n2 + 32, 19. 12 a* - 47 a^fi - 8 a'^b^ + 107 a?>3 + 56 b*.
20. 2 x2 - 8 y2 + 24 ?/2 - 18 ^2. 21. 8 a^ + 40 ac - 18 6'^ + 50 c^.

22. a5 - 6 a2 - a - 6. 23. x^ - 32. 24. m^n - nm^.

25. 10x6 - 13x* - 52x3 + 26x2 + 58x- 9.

26. 8x*'»-i?/«+* - 22x2"»+2?/3n+i + ]5xV"^-

27. 6 m5 - 13 m* + 4 to3 + 9 m2 - 11 m + 3. 28. 32 a^ + 243.

29. a^ - b a*b + 10 aW- 10 a-b^ + !ia¥-b^. 30. x^ -6x^ -3x2- 1.
31. a6-12a* + 48a2_64. 32. m&-8»n*K + 48?>i2n3 + ll TOn<-28n^

33. x6 - 6x< + 13x2 - 9. 34. a^ - 3 abc - b^ - c^

35. 12 x5 - 2 x*y - 22 x^y"^ + 9 x^j/S + 8 xy* - 4 y^.
36. x3 _ 9x2 + 26 X - 24. 37. 8 a3 + 26 a2 - 67 a + 15.

38. x6 - 2/6. 39. 60 ?i? - 127 n2 _ 214 n + 336. 40. a^ - x^.

41. 4 m* - 73 m2»i2 + 144 „4. 42. ^e _ 1. 43. x»+x*+l.

44. 4a*-13a262+9 64. 45. i6x6-144x4-a;2+9.

§ 53; pages 35, 36.

2. 11x2-111. 3. 2a. 4. 2ab-2mn. 5. -4xy + 4x«.

6. a2 + 62 + c2 + (^2 _ 2 a6 - 2 ac + 2 ad + 2 6c - 2 ft(Z - 2 C(Z.

7. 16 X* - 72 x2 + 81. 8. 2 a26 - 2 a62. 9. 4 x2. 10. 0^ + 2 a3x3 + x^.
11. a8 - 68. 12. 12 x2_+ 12. 13. - x2 - 2/2 - 22 + a^y + j/2 + 2X.
14. 0. 15. 16a3-2a. 16. 3x2 + 3y2 + 3^2 _ 2a;2/ - 22/0 - 2 2x.

17. 4 a* - 64 x«. 18. 8 6c. 19. 6 m* + 16 mhi + 16 mn^ - 6 n*.

20. - a3 - 63 - c3 + a26 + a62 + a2c + ac2 + 62c + 6c2 - 2 a6c.

21. 6a26 + 2 63. 22. - 2x3 - 24/3 - 22;3 + 6x2/«.

§ 61 ; pages 42 to 44.

8. 6x-7. 4. 5m + 4n. 5. 2a-3. 6. x^ + a; _ 12.
7. 4 m2 - 6 mn + 9 n\ 8. x2 + 4x2/ + 16 2/^- 9- 2 a - 4,

10. -10X2/-6. 11. 5a26+6a62. 12. m^-mn-Z 71^. 13. 3a + 4.
14. ia^b-ab"^. 16. a-6 + c. 16. 2x-4j/. 17. 5 jn2-3mn + 4n2.

18. 2a2-3a + 5. 19. x24-2x+l. 20. n-2. 21. 2m3-3wi2-5m-l.

22. x2+xj^ + ^2. 23. l-2a2+4a*-8a6. 24. 8x3 +12x22/ + 18x2/2 +27 2/3.
25. m2-3m-4. 26. 3x2-x-2. 27. a2+a-l. 28. 2x2 + 9x-5.

4 ALGEBRA.

29. 4m2-2 7?i?i2+n*. 30. a;i-2a:H4x2-8a; + 16. 31. 10a2+3a-4.

32. m2-l. 33. a + 3. 34. 4 x'»+5y2 _ 4 a:42/n,

35. 2a* + 2a^b + 2 a^b^ + 2 aft^. 36. a^ + a% + ab^ + ft^.

37. 2m2-3. 38. 4a2-12a + 9. 39. 2x3 + 5x2 - 8x - 7.
40. x3 - 3x2 _ 3. 41. a2 _ 2(z + 10. 42. x2 - 6x2/+ 9j/2.

43. 3 x3 _ a;2 _ 2 X - 5. 44. 2 a3 _ 5 a^ _ g ^ + 4.

45. m^ - 2 m2n - mn'^ + 2 n^. 43. 4 a 4- 6 - c 47. xp + yi - 0^
49. x-c. 50. x2+(a + 6)x + oZ>. 51. x-2 6. 52. (« + 6)x-c.
53. (m — h)x — p. 54. x + a. 55. x2 - (6 + c)x + 6c.
56. a(6-c)+d. 57. a+(2m-3n).

§ 62 ; pages 45 to 47.

2. 270. 3. -9. 4. 42. 5. 729. 6. -5. 7. -^.

15

8. - 748. 9. 854. 10. ^. 11. - -• 16. 9(x + y)^- 25.

17. 63(a - 6)2 - 20(a - 6) - 32. 18. 2(m + «) + 3.

19. (X - 2/)2 _ (X - 2/) + 1. 21. V « - tV ft + t! c.

22. _|^x + i|j/-T%0. 23. -ia_/^6 + -V-c.

24. _ Jjx - f 2/ - It^- 25. ^V«' - H- '

26. ija?-^\a% + j\ab'^-i^b^. 27. fx2-|x+2V

28. I a2 - f a6 4- i 6'^. 29. a^6* ~ 2 a2p+363s+2 + o^ft^?.

30. x^+i - x^r'^+i. 31. a2^+3 + aP+2629-i + a6<9--'. 32. 2(x + 1)2 - 3.

33. -5(x+2/)2-10x(x + 2/)4-15. 34. 8x-2. 35. fx2-fx-ii

36. x3 + (a + 6 - c)x2 + (aft -be- ca)x - abc. 37. a'^+^b^ + a^b"-^.

38. _ia2+iio_|3^ 39. („i_,i)4_2(7n_n)2+l.
40. a3''+i62 4. aft3n+2, 41. 4 ^^2. 42. 0.

43. i^e «* - f «^ - t a^ + ! « + ¥•

44. 3 ^2 _ _^ a + 1 45. a3 - 3 a26 + 3 a62 - fts.

46. 3 x2'»-i2/8 - 7 x2?/2»+i. 47. (a + 6)x2 + (a^ + 62)x - 2 a6(a + 6) .
48. (a - 6)2 - 2 c(rt - 6) + c2. 49. x-"" - x"?/" + y-\

60. t a* - I a2x2 - f aa;3 _ j. ^4.

51. x3 + (— a + 6 — c)x2 + (— a6 — 6c + ca)x + abc.

52. x2p + x-i + x2'- — 2 xP+« + 2 XP+'" — 2 x9+^

53. f x2 - 1, X + ■]. 54. x- + (a - 6)x - «6. 55. .x3 + y^ + z^ - S xyz.
66. 2 a262 + 2 6^02 + 2 c2a2 - a* - 6* - c«.

ANSWERS. 6

§ 75 ; pages 51, 52.

3. 14. 8. 2. 13. -A. 18. -6. 24. - -• 29. — •

11 7 11

4. -7. 9. -. 14. -• 19. 5. 25. 4. 30. - -.

7 5 4

5. 4. 10. --• 15. 3. 21. -5. 26. - -• 31. -1.

3 5

6. - 5. 11. 1. 16. - -• 22. 2. 27. -■ 32. — •

9 2 3

7. -9. 12. ^. 17. 8. 23. -10. 28. -6.

§ 77 ; pages 55 to 58.

5. 10,9. 6. 159,87. 7. 24,14. 8. A, f 7.50 ; B, f 5.25 ; C, f 9.25.

9. A, 65; B, 13. 10. A, 42; B, 84. 11. A, $ 12 ; B, $36.

12. 9 five-cent pieces, 7 twenty-five cent pieces. 13. 8. 14. 17.
15. 6 fifty-cent pieces, 11 dimes. 16. 47,29. 17. 9,4. 18. 13,7.
19. A, 43 ; B, 57. 20. 9 oxen, 27 cows.

21. 3 dollars, 12 dimes, 15 cents.

22. 3750 infantry, 500 cavalry, 125 artillery.

23. A, 320 ; B, 1600 ; C, 3840. 24. A, $25 ; B, $18 ; C, $40 ; D, $32.

25. Wife, $864; each son, $72; each daughter, $216.

26. A, $42; B, $23; C, $ 29 ; D, $31.

27. 13 three-penny pieces, .36 farthings. 28. 44, 27. 29. 324 sq. yd.

30. 12. 31. 35, 36, 37. 32. A, 68 ; B, 18.

33. 8 $2 bills, 13 fifty-cent pieces, 24 dimes. 34. 7, 8.

35. 3, 4, 5, 6. 36. Worked 22 days, was absent 10 days.

37. 6 bushels of first kind, 18 bushels of second kind.

38. 75 men on a side at first ; whole number of men, 5668.

39. First class, 75 ; second, 115 ; third, 150 ; fourth, 195.
40. 18. 41. A, 8 minutes ; B, 5 minutes.

42. 15 pounds of first kind, 35 pounds of second kind.

§ 82 ; pages 60, 61.

25. a2 -h 2 ac + c2 - 62. 30. 1 _ ^2 _ 2 a6 - 62.

26. a;2 _ 2xy + y^ - z^. 31. x* - 2 x^ + 1.

27. a2 _ 62 _ 2 6c - c\ 32. a2 - 4 62 -|- 12 6c - 9 c^.

28. a< - a2 + 2 a - 1. 33. a* + a^lP' + h^.

29. x*-5a;^-f-4. 34. 9x2 - 16j/2 - I62/2 - 4g2

6 ALGEBRA.

§ 99 ; pages 72, 73,

38. (a-b-\-c)(a-b-c). 44. (3a-4 6 + 2c)(3a-4 6-2c)-

39. (m + n+p){m + n-p). 45. (4:X+2y- 5z)(ix-2y + b z).

40. (a-{- X + y)(a — X — y). 46. (m — 2n + x){m — 2n ~ x),

41. (x + y-z)(x-y + z). 47. (2 a + 6 + 3) (2 a - 6 - 3).

42. (a + 6 + 2)(a + 6-2). 48. (5x + 2/ +32;)(5x + ?/ - 3^;).

43. (1 + ?n - n)(l - m + ?i). 49. (a-6 + c-(Z)(a-6-c + d).

50. (a + X + & - j/)(a + X - 6+ y).

51. (x — m + y + n) (x — m — y — n).

52. {X + y + a + b){x + y - a - h).

53. (2 a + 6 + 3 c - 2) (2 a + 6 - 3 c + 2).

54. (x - 4 2/ + 2 + 6) (x - 4 2/ - ^ - 6).

65. (5 a — m + & — 3 n) (5 a — m — 6 + 3 n).

§ 106 ; pages 78 to 80. '

30. (1 + n)2(l - ri)2. 45. (2x + 3y)2(2x - 3y)2.

41. (a + 3)2(0-3)8. 46. (a-l)2(a2 + a + l)2.

42. (x+ l)(x-2)(x2 + x + 2). 52. (3 a + 2)2(3 a - 2)2.

43. (a + 2 6X«-2&)(c+3(?)(c-3d). 53. (x - 2)(x + 3)(x -3)(x + 4)

44. (x + l)(x-l)(x-4)(x-6). 54. (a - 1)*.
55. (a - X) (6 + y) (a^ + ax + x^) (62 - by + y"-).

57. (6a + 26-7c)(6a-26 + 7c). 59. (x + l)2(x + 2)2.

60. (a + l)(a - 2)(a2 - a + l)(a2 + 2a + 4).

63. 2 6c(a + 6 + c) (a - 6 - c).

64. (a -1) (a + 3) (a + 4) (a + 8). 66. (x - l)(x + 2)2(x - 3).
67. (a + 6.+ c)(a- h + c){a + ^ -c)(a - 6 - c).

76. (m + x)(m2-4mx + 7x2). 77. 6(3a2 - 3a6 + 62).

78. (x-2/)(9x + y). 79. (a + 6)(a2 - 3»a6 + 62).

80. (a + 6 + c + d)(a + 6-c-d). 81. 2x(x2 + 3).

82. (x + 2/)(2x2 + ?/2). 83. (a + l)2(a - l)2(a2 + 1).

84. (a + X + 6 - 2/) (a + X - 6 + ?/) . 85. m(x - m) (m - 3 x).

86.2 2/(3x2 + 2/2). 88. (x + l)2(x - l)(x2 + l)(x2 - x + 1).

89. 3a(a-l). 90. 7(5m - l)(??i2 - m + 1).

91. (x + y - z)(x-y + z){x + y + z){x -y - z).

92. (o-56+4c + 3d)(a-56-4c-3d). 93. (l + a)C3-a-a2).

ANSWERS. 7

§ 117 ; page 89.

6. a; - 1. 6. 2 a + 3. 1. x + 2. 8. x - 3. 9. m + 1. 10. 3 a - b.

11. 3 a2 4. ax -2x2, 12. x(2x-5). 13. Sx + iy.
14. 2 a3 - 3 a2 _ « + 4. 15. 2 m2 - mn + n"-. 16. x - 2.
17. a2 + 2a + 4. 18. jm^ - 2 mx - 3 x2. 19. a - 1. 20. m2(m + 2).
21. a -66. 22. x + 3. 23. 3a2-2. 24. a + 4. 25. 2x-i/.
26. 2x2-3x-l. 27. x - 2. 28. ax{a + x).

§ 118 ; page 90.
2. 2x-9. 3. 4a + l. 4. 3m + 4. 5. 5a -2 6. 6. x + 2.

7. 0+ 1. 8. m- 1, 9. 2x-3y.

§ 125 ; page 93.

30. {x ■\- y -\- z)(x - y -V z)(x - y - z). 40. (m + w)2 (m - n)2.

41. (a + 6 + c) (a - 6 - c) (a + 6 - c).

§ 126 ; pages 94, 95.

2. (2x + 7)(2x2- 19X + 45). 5. xy(6x - 2/)(8x2'+ 21 xj/ + 10y2).

3. (a-4)(3a2 + 14a-5). 6. 3(4 jw + 5)(4?ji8-ll m2-6m + 9).

4. (3o+86)(12a2+16a6-362). 7. (2 a + 3)(3a3 - 14a2 - a + 6).

8. x(2 a2 - ax + 3 x2) (2 a^ + 5 a2x + 2 ax2 - x^).

9. (2 a - 3 6) (a* + a86 - 5 a262 + 2 a63 + 6*) .

10. (3x-2)(4x*-5x2 + 4x-3). 11. {ai-Za + 2){'^a^-Qa-^).

12. 2 mn(3 m'^-mn-2 n"^) (3 m^ - 2 mH - 7 mrfi - 2 n^).

13. a2(a2 _ 2 a + 3) (3 a* + 11 a3 - 6 a2 - 7 a + 4).

14. (x2-x-3)(3x4 + 7x3 + 6x2-2x-4).

§ 127 ; page 95.

1. 8x* + 20x3- 46x2 - 117x- 45.

2. 162a* + 117a3- 147a2-62a + 40.

3. 12 m*- 10m3-86?n2 4. i4o,„_48.

4. 24 x^ - 70 x6 - 15x5 + 25.x* + 6 x^.
6. a^ + 2 a* _ 10 a^ - 20 a2 + 9 a + 18.

§ 133 ; pages 98, 99.

12. 3a. 13. 2^. 14. ^^±1. 15. ^('^-^). 16 ^« + ^^
46 5 2/2 a-1 x-6 5a-26

8 ALGEBRA.

j^ m - 8 jg X + y jg a(8ffl + 7x)_ g^ x — 9m

m(m — Q) ' 2xy x(8a — 7x) * x + 3m

21 «H2ffl + 4 22 2m-5 23 £±^±5. 24 9 a'' -12 aft + 16 fe^

a2+l " " 3m+i ' x+y-z ' Sa + ib

25 1. 26. a + b + c + d^ 27. ^^ + ^. 29. ^ + .'^. 30 ^^-.
a-6 + c-d 3x-2 4-m 7-2x

31. 6y-x^ 32. ft-«. 33. 9«-l 34. « + ft + c.

?/+x d+2c 4a-^ + 2a + l a — b + c

§ 134 ; page 100.

2 x+3 g a-2 ^ 2x + 5y g 2m-3 g x-2
6x+7* ■ 2a-l* ■ 2x-9y' ' 3m+4' ' x2-3x+l'
^ 3a + 2& g 3x-2 g 2a + 1

4a-6 ' ■ 3x+ 1' ■ 6a-l"

-Q ?»2 — OT + 3 ,, g^ + 3 ax + x'^

' to2 ^. 4 ^ _ 2 ' a2 - 2 ax - 4x2

§ 137 ; page 102.

6. 4x-6+—^ — 11. 3a_2«-t-5^

2x + 3 4a-l

6. x2 + xy + j/2 + ^l!_. 12. 3m2 + 4 --lil±A.

X — «/ 4 m2 + 1

7. a2 - a6 + 62 _ A^. 13. x^ - x^y + xy"^ - y^ + ^^•

a + b x + y

8. 5a2-3a-l 14. 6a + 7 ^a-^

3a + 4 3a2-4a + 5

9. 2 m + 5 n + ^^ "^ — 15. a* + a36 + a262 + a63^.;,4+ 2^.

2 m — 6 w a — 0

10. 2X-1+ '^"^ — 16. 4x2 + 6x-2 ^^~^ ■

X2-X-1 2x2 + x-3

§ 138 ; page 103.

- 3a2-lla + 2 ^ 2a: _ 10a2-13a-9 „ 4x2-10x-7

8. • 4. • 5. ■ • 0.

3a x—y 2a— 3 6x

7 2& g 7>i^ — n^ g 10 X iQ 18x2 -J 4 3-j/

3a + 6 m + n 2a — 5x '3x — 4 x + 2y

j2 8m8+24?n2-36 9ra-27 jg 4a3+5a j^ 5a2_23a^ + 8&^

2m+3 * ■ 2a-l * ' 4a-36

^^ 2x2y+2xy2 ^^ a*+ft^ j^ -3x2 jg 54 n^

x2+x2/ + y2 a — b x2+x+l m" — 3mn + ^.)n'^'

ANSWERS.

§ 140 ; page 105.

g 9x^-Sx 10x2 ^ 2a^-2a'^b + 2ab'^ 4a&2-4 6S

' 2a;(9a;2-l)' 2x(9a;2_i)' * (^a-b)(a^+b'^) '{a-b^ia^+t

9.

3(a-l)(«2 + l) 6(a + l)(a2 + l) 9(a2 - 1)
a*-l ' a* - 1 ' a* - 1 ■

10 2 a:3 - 16 .Sx^ + 6x* + 12x3 3 x^ - 12 x*

3x2(x-4)(x3-8)' 3x2(x-4)(x3-8)' 3x2(x - 4)(x3 - 8)'

12.

jj x2 - y2 (ffl - by

(a_6)(a;_y)2' (a_6)(x-y)2

(a+5)2 a2-9 a2-4

(a+2)(a-3)(a+5)' (a + 2)(a-3)(a + 5)' (a + 2)(a-3)(a + 5)'

§ 142 ; pages 106 to 111.

g 21a -4 ^ 20x2- 18 y g 6x + 1 g 3a2_i4a;2

24 ' ■ 15xV ' ■ 48 " ■ 18rt2x2

. - 4x2 + 3m2 g ab + bc+ eg g 10 a - 53

96 mx abc 28

jQ 20x3-4x2 + 57 x+35 jj 4bc-9ca+8ab ^^ 3x-10y

40x3 " ■ 24 a&c * ' 30x

.13. ^. 14. 1. 15. ^^-^ 18 ^^a-9

36* ■ 20" ■ 108 ' ■ c3a + 5)(4a-7)

19 «^H1 20 7X-22 2j o^+fo^, 22 «^-15a + 3

m2-l* ■ (2x+l)(5x-6)" ■ a2-62 ' a2-3a-28

23 2m2 + 2n2 ^ 4x . 25. _iiL_.

■ m2 - n2 x2 - 1 4 a2 - 1

26 2x 27 10^^^

x-2^' ' (2a + 3 6)2(2a-3 6)

28. :^^^ 29. l^'^^

(x-2)(x + 6)(x-9)' ■ (x-3a)2(x + 7«)

30.-^. 31. -i^. 32.1. 33.-2^. 34.2^^=2x

a + b
4x2 oo 12a + 18

(x+2)=

35. -

39. ^"^ . 40. 0. 41

8a3+63

44. 0. 45.

**• o2-a+l' ^^' ^' ""■ Cl+x)(2-x)(3+xy ""■ xy{x-y)

XO ALGEBRA.

12(x-2) a2-9 m(16-»i2) I -a

59. _^!_. 60. ^-^. 61. — ^ 62. -2«i_.

a;2 - 4 a; + y 9 - 4 a2 ni + 2

63. 64. 1±1^. 65. 68. 0.

(a + c)(6 + c) l-a;3 (a; - y) (2/ - «)

§ 144 ; pages 112, 113.

4. ^. 6. I. 6. 2abc. 7. ^^. 8. |^.

5(q+6) jp 3to + 1 u 2a:2(x-3) jg y(x + 2y)

3(a + l)* ■ m-5* * (x-6)2 ' a;(x + ?/)

(a-46)(a-25) ^^ xCx"- - 1) j^^ 1. ^g, -^.

a(a-36) ' (x+2)(x2 + x + l) 2 a-1

2x-3y jg (x + y-g)^ 19 (a-b)\ go 1 21. ^±^.
x-y ' {x-y-zY {a + by^ 2x

9.
13.
17.

§ 146 ; pages 114, 115.

3 3a2 ^ 9m^^ g 3(x + 3) g m(2m + 5w)
7 6xV* ' 4rt* * ■ 2(x-2)" ■ n(4?)i-3n)

- 3(2a-5&) 8 a{a-\- 7) g x(x + 2 y) ^^ _^.

5(4a + 3 6)' ■ (a -3)2" ' y ' x'^ - 1

jj a(«-2) 12 Cot + 2&)(a-5&) jg 2fl! + x i^^ a + & + c

0 + 5' ' (a + 86)(a + 4&) ' a + 2x * a-h + c

§ 148 ; pages 115 to 117.

4 2 g X-+X + 1 g_ m + n. 7. 2x-3y
' 2m-l * X w-n

9 a^ + 6y. 10. «. n. ^±11. 12.

x + 2y h y

13. ^izj. 14. ^izii^. 15. a + 1.

X + ?/ a^ — b^

jg 103X + 78 ig 2 - 3 g go ^-±-i^.
39X + 30* '5-7 a' 'Sa-fe

22. — ± 23. ^i±-X^. 24.

2Cx + y) 24 n(?>t- w)

6

(X-

3)(x

+ 2)

X

16.

2a2-

-3 62

7

ab

21

2(x

-a)

X + «

25. -

ab

1+x^' ' (x- yy ' m ' a^ + b^

§ 149 ; pages 117 to 119.
79a-31 o a^^ <* ^x

1 tvit—ai 2 «"0" 3 ___fj£__. 4,

4a+3 * ■ a^-ab + b^ ' (2a-3x)8 " 1 + x + y+xy

ANSWERS. 11

5. ^^izil. 6. ^^--3y. 7. «^^:i^. 8. 0. 9. L_.

a2 + 1 2 x?/ a'W x + 2y

10. (a:-2)(x-8). ^^ ^^ 12. 4a2_9. 13. ^x^

(a;-l)(x + 5) (x-2/)(x3+J/»)

6(a2-a6 + 62)' " xV " ' ac-bd ' (x-8)(3a;-8)'

18. 1. 19. »^(2^^ + n3). 20. ^^±^. 21. —1— 22. 2.

n2(m2 - n2) a - 6 1 + x^

23 (^-b)\ 24. ^^'-5-'^ + ^. 25. 2 26. 1.

2(a + &) 3x2-4x-:i a(x + 2a)

27. ^" ~ ^^ 28. x2 + 1. 29. ?H + 2 ?i. 30. a + b - c ^

X- + 2/- «6(a — c) (c — 6)

gj_ X- + xj/ + ?/'^ 32 2(x + y) 33_ a - &.

x2 — xy + 2/2 ' (x — z)(y — z) a + b

34 8 gg SgS 3g 6x2- 12 3^ 2b(a+b)

l-x8* ■ a8-256" ■ (x2-l)(x2-4)' " (a-6)(a2+62)"
38 2(x2-l) 3g 20x2-34

■ x* + x2 + l" ■ (3x-l)(2x + 5)(4x + 3)

13 a

40. -

(2a-3)(3a + 4)(5a-2)
§ 151 ; pages 120 to 124.

2.

10.

13.

8
3"

23.

1
4'

35.

-4.

45.

4

5"

3.

-2.

14.

-5.

24.

2.

36.

4.

46.

-2.

4.

_3
2"

15.

4

3"

25.

5_
2

37.

11
6*

47.

1_
2'

5.

3
5'

16.

2
3*

26.

2

7"

38.

7
3'

48.

_9
2

6.

5

7"

17.

6.

27.

3
5' •

39.

3

5'

49.

1 ^
11'

7.

1
2

18.

-3.

30.

11
4'

40.

_ 2_
17'

50.

19
3'

8.
9.

4.
5

s"

19.
20.

_1

2

7.

31.
32.

-1.
4
5*

41.
42.

2.
19
9*

51.
52.

2
3"
5.

10.

4
3'

21.

-1.

33.

1
3*

43.

43

"7"

53.

2
5'

11.

- 1.

22.

-4.

34.

1

44.

3.

L2

ALGEBRA.

§ 153; pages 125

126.

2.

2 a

3 ft"

8.

-2 a.

14.

2
n

20.

a + b.

3.

he
a

9.

1

a — b

15.

ac
~b'

21.

a + b
2

4.

— a.

10.

2(a-6).

16.

-3 a.

22.

3mn

m + )i

5.

Zn

11.

m + n.

17.

62

a

23.

aft
a — ft

6.

a-\.

12.

2a + 6
2

18.

2a
Sb

24.

a + 5
2

7.

m — 1
m

13.

12(a-6).

19.

2a -3b.

25.

-ft.

§ 154; page 126.

2.

.09.

4. 5. 6.

93

500 '

8.

.6.

10. 0

3.

-4.

5.

- 20. 7.

-.02

9.

-1.4.

18. 48.

19. 82.

20.

79.

23. -.

8

26. 35,

14.

28. 107, 27.

29.

22 _

is'

riage, f 175.

32.

6.

§ 155; pages 127 to 135.

2. 40. 3. 56. 4. 42. 5. 27, 18. 6. 32, 24.

7. A, $40 ; B, $48 ; C, $ 36. 8. Water, 288 ; rail, 360 ; carriage, 120.

9. A, 24; B, 64. 10. $25. 11. $2.45. 14. lOf. 15. I3V

16. 15| hours. 17. If minutes.

21. 20. 22. A, 24; B, 48.

27. A, 30 miles ; B, 36 miles.

30. 59. 31. Horse, $ 250 ; ca

33. Horse, $ 180 ; carriage, $ 280 ; harness, $ 30.

34. Express train, 45 miles an hour ; slow train, 30 miles an hour.

35. A, 32 miles ; B, 25 miles. 36. 120.

38. 38j-2j- minutes after 1. 39. 38523- minutes after 6.

40. 21 j3j- minutes after 4. 41. 10}^ minutes after 5.

42. 87. 43. 22J miles. 44. A, 3 days ; B, 6 days ; C, 8 days.

45. 49xV minutes after 9. 46. A, $36 ; B, $32 ; C, $27.

47. lOia minutes after 8. 48. 45 minutes. 49. A, $1200 ; B, $900

50. Longer piece, 30 yards ; shorter, 24 yards. 61. $ 1840.

52. 21j8j and ^i^j minutes after 7. 63. 9/^ minutes after 2.

ANSWERS. 13

54. Gold, 1540 oz. ; silver, 420 oz. 55. $4725.

56. A, 4; B, 5; C, 6. 57. 2 p.m.

58. $ 1250 in 4.\ per cent bonds, f 1750 in 3^ per cent bonds.

59. 24 miles an hour. 60. \Gj\ minutes after 10. 61. 7.
62. $18000. 63. .$2400. 64. Gold, 57 oz. ; silver, 70 oz.
65. Fox, 180 ; hound, 13.5. 66. $5400.

§ 156 ; pages 136, 137.

071 am „ . am — amn „ a — an

2. , 3. A, years ; B, years.

m -^ n 971 + n m — n m — n

4. -miL_. 5. ^ 6. " - ^^'' dollars. 7. ^- + ^.

m + « db -{■ he + ca \ + r a — a

8. J?^ miles. 9. ^^. 10. ^^"^ dollars. 11. ^^^(^ ' P\
b + c b - a 100 + rt pr

j2_ lOO(a-p) per cent. 13. «^±^, ±

pt ft + 1 ft + 1

14. A, S^ miles ; B, -«?L_ miles. 15. «^^ + ^" + 'P cents.

??» + n 77i + n a + 6 + c

16. Fir.st kind, ^^'^^ ~ ^^ ; second kind, ^'^^ ~ ^.
b — a b — a

17.

a?i

1 + w + m7i 1 + ?i + m,n 1 + n + m?i

. 2mnp _ 2 mji;) _, 2 7?uijj

18. A, ; B, ; C,

wyi + 7ip — >?ip ?nj) + 7ip — mn mn + mp — np

§ 164; page 141.

3.

x = 2.

2/ = 3.

8.

x=l.

2

11.

X = —
2

2.

14.

x = -l
2

4.

X = -'4.

-1-

"=6-

15.

^ 2
x = 4.

5.

X = 3.

2/ = -5.

9.

— 1

12.

-I-

2/=-l.

-

16.

x = - 6.

6.

X =— 1.

y = -4.

-1

2/=-3..

—I-

4

17.

x = - 3.

7.

x = -§.
4

3

10.

13.

X = —

5.

18.

y = 5.
x = 9.

^ = 4-

"=6,-

2/ = 4.

2/ = 7.

14 ALGEBRA.

§ 165 ; page 142.

2.

3.

x = 3.

2/ = 4.

x = -4.

8.

x = l.
^ 3

11.

x = —

2/=^-
^ 2

2.

14.

4

2/ =-4.

4.

y=-i.

x = 2.

9.

x = -t

12.

x = ?.

15.

X = — 5.

y = e.

5

2

2/ = l.

5.

X = 5.

y=-7.

y=-l-

-1-

16.

^-\

6.

x=-l.
y = s.

10.

x = -h
4

13.

3

2/ =-3.

7.

x=-3.
y=-2.

y = '-

^ 3

2/=-

3^

17.

x = 4.
y =- 6.

2.

a; = 2.

2/ = 5.

3.

a: = 4.'

2/=-3.

4.

a; = -5.

2/ =-4.

5.

x = \.

y=-2.

6.

x^-2.

-1

§ 166; page 143.

7.

x = -^-

10.

x = l.

13.

X = 6.

4

3

y=-l.

--I,

^ 2

14.

x = -\.
2/ =-5.

,Q

15.

x = 3.

2

11.

x = -.

8.

" = 3-

2

y = 2.

^=r

2/ = 3.

16.

x = -l.

5

5

12.

a: = -l
3

^ 2

9.

a; =-3.

2/=-'-

17.

x = -4.

2/ = l.

4

2/ = 7.

167 ; pages 144 to 146.

2.

a; = 6.

6.

x = -8,

2/ = - 10.

2/ = 5.

3.

x = 12.

7.

x=-6.

2/ = - 12.

y=-3.

4.

x = -l.

8.

x = 3.

2/ =-5.

2/ =-5.

5.

x = 4.

9.

x = 18.

y=-3.

2/ = 6.

10.

x = 4.

14.

X = — 5.

2/ =- 5.

2/ =-7.

11.

x = l.

15.

x = -7.

2/ =-2.

2/ = 8.

12.

x = -l.

-1-

2/ = 5.

16.

13.

x==5.
y =9.

— 1-

ANSWERS. 15

17. . = -12. 20. x = .S 23 ^^44 ^^ ^^11.
y=-G. y = -.01. 5 2

18. x = 5. 21. x = 2. 2/ =-11. w=-?

6 5 ^2

y = — y = •

2 ''3 24. x = 7.

19. x = - 2. 22. X = 3. y = 10. 26. x = - 10.
»/=-6. ?/=-l. 2/ = 5.

§ 168 ; pages 147, 148.

2 ^ ^ 35 g + 24 fi 7. x=-2a. 14. j. ^ ^j,

23 y= 6. y _ ^^

^, 14 a -18 6 o ^

23

y=-2n. y = a.

3. a:= "+^ ■ q „_aa^(6c^+_6^

g_ ^^aa\oc' + 0'c) 16. x = m2n.

_ dwi + bn — ^

2.

X =-3.

y=5.

3.

x = l
4

— 1

4.

x = 4.

2/ =-6.

C(ffi

a2

+ 62

c(a

— m'n

mn'

— np' '

— 7n'n

y — mn'^.

^'^"^^ '= cc'la'b + ab') 11. x---^.

a

4. x= ^^'P-^^P' 10. x=a. «_6

mn' — m'n ?/=—&. ^~ — t] —

_ mp' — m'/)

^ ~ mn' - m'n 11. a;=^- 18. x = ^-±-?-^.

2

5. x = ""'-^'^". y=-i- ,._ffl-26

ad + 6c 2 y - — ^

_ cm — aw 12. x=a2+6.

acZ + 6c' ?/ = a-62. 19. x = ^-^'

2

6. x = a + 6. 13. x = a(2a + 6). _a-6
y = a-b. y = b(a + 2b). ^ ~ ~2

§ 169 ; page 149.

5. x=-^i±^. 7. x = -6.

y=-2.

y = ^ 1^ . 8. X = 3.

y = 4.

6. x = '"" ~'""- 9 a.^«.

6

2,=zii: ^z^. y=-i

m'p — mp^ a

16 ALGEBRA.

10. x = a+b. 11. X =-

2'

a + b ^ 3

§ 170 ; pages 151 to 153.

3.

a; = 3.

y = 2.

z=-l.

10.

X = 1.

— 1-

17.

x = -?.
5

1

22.

« = 6.

x=-7.

4.

x = -5.

2/ =- 4.

2=2.

11.

— i-

x = ~3.
2/ = 4.

2=1.

3

23.

y = s.

2 =-9.

X- -A,

6.

x = 2.

2/ = 5.

.=1

2

18.

x = 2.
2/ = 4.

6 + c"

y- ^.

z^-1.

12.

2/ = 4.

2 = 6.

c+a

9

2= — .

6.

X = — 4.

2/ =-5.

2 =-3.

19.

x = l
4

a '-b

2 =-6.

13.

a;=- 1.
2/ = 6.

»=!•

24.

u = — b.

7.

x = -6.

z=-4.

2

-1-

x = 4.'

y=-7.

14.

X = 5.

2/ =-3.

z = S.

2/ = l.
2=3.

20.

X = a.

2 =-2.

8.

X=:-2.

15.

x = - 3.

2/ = — «-•

25.

?« = 10.

y =-5.

2 =-8.

2/ =- 5.

2=- 7.

z — — a^.

X = 2.

2/ = 4.

9.

16

16.

x = 2.
3

21.

x = ^.
a

2 = 6.

— i-

2/=-'.
^ 4

•^ b

26.

x = 6. .

z=i.

.=-1

z = -±

2/ =-2.

4

6

c

2 =-4.

27.

a; = 2.

28. x = 6.

29.

X

= -

12.

2/ = 3.

2/ = 14.

2/

= -

24.

2=- 1.

2 =-12.

2

= 36

30.

^_ 2a6c
afe + ac— 6c

„_ 2ff?)c

2

2rt6c

ab + bc—ac

ac

-\-bc-ab

31. X = ab.

y ~ be.
z — ca.

32. x =

2 be

b + c-

a

2ca

c + a —

b

2ab

ANSWERS.

33.

X — a.

z^h
a

34.

x = 3.

2/=-l.

2=5.

17

a + b

§ 172; pages 155 to 164.
3. 35, 24. 4. 20, 12. 5. — • 6. — • 7. Apples, $1; flour, $3

8. A, 24 ; B, 40. 9. 26, 15. 10. — •

16

11. A, 35 ; B, 27. 12. A, 15 ; B, 22i.

13. .^630 in 4.} per cent stock, .$ 810 in 3^ per cent stock.

14. Income tax, $28 ; assessed tax, $36. 15. A, $60 ; B, $52.
16. $ 1.75, $ 1.50. 17. 13, 17, 19 19. 84, at 2\ cents each.

20. 45 cents ; 15 oranges.

21. '^''^'^ + ^^ persons ; each received ^^^^"^ + "^ dollars.

bm — an bm — an

22. 21 quarter-dollars, 13 dimes.

23. 26^ of first kind, 43 1 of second kind.

24. 45 of first kind, 63 of second kind. 25. A, 15 ; B, 30 ; C, 60.
26. 32 for, 22 against. 28. 97. 29. 896. 30. 83. 31. 59.
32. 4 from the first, 3 from the second. 33. 85 ft., 64 ft.
34. A, 9 ; B, 5. 35. 46?.

36. Express train, 45 miles an hour ; slow train, 27 miles an hour.

37. A, $72; B, $81; C, $63; D, $180. 38. First, 38; second, 18.

40. Rate of crew in still water, ""• + ^"^ miles an hour ; of current,

, '2 mn

TllzJ^ miles an hour.
2 mn

41. Going, 10^ miles an hour ; returning, 4J miles an hour.

42. 78. 43. 369. 44. 75 ft., 54 ft. ' 45. $375, at 4 per cent.
4g bm - an ^^^{^^^^ ^^ im(a - b) ^^^ ^^^^_ ^^ ^^ 15 . ^^ 21.

m '—- n bm — an

48. $ 2000, at 6 per cent.

18 ALGEBRA.

80. Rate before accident, 36 miles an hour ; distance to B from point

of detention, 90 miles. 51. 647.

52. A, $6; B, $12; C, $8; D, $20. 53. A, $13; B, $7; C, $4.

54. Fore-wheel, 9 feet ; hind-wheel, 15 feet.

55. A, days; B, days; C, days.

mn + np — mp mp + np — mn mn+mp — np

56. A, 8 ; B, 12 ; C, 24.

57. First, $15000 at H per cent; second, $18000 at 3i per cent;

third, $ 13000 at 5.i per cent.

58. A, — hours ; B, -^^ hours.

b + c — a a — b

59. Rate of crew in still water, 9 miles an hour ; of current, 5 miles

an hour. 60. Principal, $5000; time, 3 years.

61. A, $55; B, $19; C, $7. 62. 12, each paid $3.

63. Express train, 40 miles an hour ; slow train, 25 miles an hour.

64. A, 18 ; B, 15. 65. 3 quarter-dollars, 8 dimes, 9 half-dimes.
66. 30 of 3^ per cent stock, 20 of 4 per cent stock. 67. A, 8 ; B, 7.

§184;

pages 168,

169

3.

x<3. 4. X

>!■

5. x<|

6.

x>8.

7. ;

'<T--

8.

x>a — b.

9.

X

<1, 2/<4.

10.

x>3,

2/<2.

11.

X > 5 and < 9.

12.

7,

13. 18 01

• 19.

14.

38, 39,

, or 40.

§ 187 ; page 172.

4. x* + 4x3 + 6x2 + 4x+ 1. 6. 4a* - 4a3 + 17a2 - 8a + 16,

7. 25x*-30x3-x2-t-6x+l. 8. 9 xH24x3+28x2+ 16x4-4.

9. 36 n^ + 12 n* -60 71^ + 71- -10 n +25.

11. a* - 8 a^b + 22 a^b'^ - 24 ab^ + 9 b*.

12. 4 X* + 12 x^y + 13 x^^ + 6xy^ + j/*.

13. x6 -f- 12x5 + 36x*- 14x^-84x2 + 49.

14. 16 flS _ 40 a^x^ + a*x^ + 30 a^x^ + 9 x^.

17. x6-2x5-x* + 6x3-3x2-4x-t-4.

18. a6 + 4 a5 - 2 a* - 20 a3 - 7 a2 + 24 a + 16.

19. 4x6-20x&-t-41x* -52x3 + 46x2-24x-t-9.

ANSWERS. 19

§ 188 ; page 173.

4. x« + 6a;2+ 12x4-8. S. 21G a^- 108 a'^b + IS ab^-b^

5. 27a3-27a2 + 9a- 1. 9. \2bx^ + l50x'^y+e0xifi + 8y^.

6. m3 - 12 m2n+ 48 mn2- 64 n3. 10. G^ m^- lU m'hi^+ 108 mn^- 27 n^.

7. x6 + 15x* + 75x2 + 125. 11.27x6-135x5 + 225x^-125x3.

12. 64 x^ + 240 x«yz^ + 300 x*?/2^'= + 125 y^^z^.

13. 8 x3 - 84 x5 + 294 x^ - 343 x^.

14. 125 ai8 + 450 aisfts + 640 a^fcio + 216 ftis.

16. a8 + 68 _ c3 + 3 a26 - 3 a2c + 3 fc2a _ 3 ftSg + 3 c^a + 3 c26 - 6 a&c.

17. x6 + 3x5 + 6x*+ 7x3 + 6x2 + 3x + 1.

18. x^-y^ + 8 z^-Sx^y +6 x^z + Sy'^x + e y'^z + 12 z^x-l2 z^y -12 xyz.*

19. a6 - 9 a^ + 24 a* - 9 a^ - 24 a2 _ 9 a _ 1.

20. 8x6 + 12 x5 - 30 X* - 35 x3 + 45 x2 + 27 x - 27.

21. 27 - 108 X + 171 x2 - 136 x^ + 57 x^ - 12 x^ + x6.

§ 193 ; pag-e 176.

24. 56. 25. 135. 26. 252. 27. 432. 28. 588

29. 24. 30. 105 a&c. 31. 402. 32. 45. 33. 12.

34. 6. 35, 126. 36. 28. 37. a^ + 4 a^ + a ~ 6.

§ 195 ; pages 178, 179.

3. 2x2 + x + l. 10. 3x + 5y-iz. jg ^ + 4 _ 1.

4. l-3a + a2. ^ 7m2-mn-4n2. , ' ^ "^ ^

5. 3x2 — 4 X — 2. 17. 1 — X + x2 — x3.
e" 2x2 + 5x-7. ^^- 3a^-5a + 4. ^^ x3-4x^-2x-3.

7. a-6-c. 13. 5x2-2x^-32/2. ^^ .^ _y _3y^,

8. 2a3 + 3a2-l. 14- 4TO2 + mx2-3x^ ^ |^

20 — — _ 4- -.

9. x3 - 2 xa2 + 5 a^. 15. 3 a2 - 2 aft - 5 52. 3 2 5

21. 2a3 + 3a26 + 4a&2_563. 26. 1 + a _ ^'^ + ^+ ...

2 2

22. H-' + «^ + ^'. 27. l-^-5:_^'_...
2 3 4 2 8 16 ■

23. 3x3-2x2u-xw2 + 4v3. 28. l-^_-i^_2I«!

2 8 16 ■

24. ^ + ^-^'. 29. x + 5__9_ + ^+....
3 a a2 X 2 x*^ 2x5

25. 1 + 2x-2x2 + 4x3+ .... 30. 2a-- ^ ^

2 a 16 a3 64 a^

20 ALGEBRA.

§ 199 ; pages 182, 183.

1. 65. 10. 3581. 20. 3.6055. 30. .8660.

2. 148. 11. 274.9. 21. 6.9282. 31. .7453.

3. 713. 12. .4027. 22. 8.0436. 32. 1.148.

4. 8.07. 13. 51.64. 23. .44721. 33. .7071.

5. .396. 14. .07906. ' 24. .23664. 34. .7745.

6. .254. 15. 9.318. 25. .62449. 35. .9354.

7. 62.9. 17. 2.6457. 26. .094868. 36. .6373.

8. 9.82. 18. 2.8284. 27. .027202. 37. 1.035.

9. .0567. 19. 3.1622. 28. 2.9265. 38. 1.258.

39. .6085.

§ 201 ; pag-es 185, 186.

7. X- -2x-l. 11. a- - 3 a - 2. 14. x"^ + 2xy + 4y'^.

8. 2a^ + 3« + l. 12. 2a;2_5a; + 2. 15. ^-1+:*

9. Sif + y- 2. 13. 3 a2 - 2 a& + 62.

§ 206 ; pages 189, 190.

3 X

1.

27.

6.

9.5.

11.

.0481.

16.

1.442.

21.

.741.3.

2.

53.

7.

.608.

12.

92.4.

17.

1.912.

22.

.7031.

3.

3.9.

8.

3.59.

13.

7.63.

18.

2.087.

23.

.7368.

4.

.85.

9.

806.

14.

697.

19.

.2714.

5.

136.

10.

57.2.

15.

.1048.

20.

.8549.

§ 207 ; pag-e 190.

1. 3 a + 2 ?;-. 2. 1 - 3 x - x-. 3. 2a^-a- 2. 4. x^ + ?/2.

5. a - 2. 6. 21.4. 7. .46.

§ 217 ; pages 195, 196.

8. mT. 9. c^\ 10. Gn'K 11. 7a~i». 12. Q ab\ 15. ax~K
17. a-h. 18. 8X-2 + 27. 19. 8 a-2-18 a-i-47-15a. 20. x-3-16.

1112 T fi _4 _5 t 3 _"

21. x^+x^y^ + yK 22. m^7i~^ — im^n ^+6m7i ^ —4m^7i~- + m'^n ^.
23. a-V)-^ - 3 a-'^b-' + a''b-^. 24. 2 m~^ + 4 m'^n'- + 18 ?r*.

25. 4 ah-^ -11 ah- + 16 a~h^. 26. I8mix~^-20m^x^ + 2m~^x.

ANSWERS. 21

§ 218 ; pages 196, 197.

5. 6^. 6. 2xM. 7. «l 9. 3x~^'>. 11. a^ - ah^ +b\

12. a~f + (f- + a~^ + 1. 13. x- - 2 + x"-. 14. aS -2 a^ + 1.

15. x-2x''-y^-\-y^. 16. m-'--2m-^ + l-2m. 17. Sx-V+a^-JZ + a^-
18. a-m-'^ + ffl3„i-2 - 2 a^jjrs. 19. a^b"^ - 2 - 3 oThK

_3 4 2 3

20. tii^x ^ + 2m' + m'^x^.

§ 220 ; pag-e 198.
8. X*. 11. cfi. 13. cl 15. m"i

10. mi 12. x"i 14. a"^. 16. x".

§ 221 ; page 198.

2. 125. 6. -. 10. — • 14. — .

7 32 64

3. 24.3. 7. - -• 11. - 128. 15. - 1024.

3

4. 256. 8. 128. 12. 32. 16. 81.

5. 27. 9. 49. 13. 625. 17. - -•

8

§ 223 ; pages 199, 200.
8. 2a* + l-5«~^. 9. 3x~3_2x"3 + l. 10. ah-'^-^ah'^-^ah-'^.

16. x~2 + 2x^-3x1 17. a*'"'. 18. a'-'-S". 19. x"'-i.

20. x"'-!. 21. x""". 22. a'. 23. -• 24. x^""'. 25. ^-^t^-

8 1 - a

2e_ 2(^ + 2^). 27. a" + l + a-". 28. ^x-^^Cx+y),

29. x" + 2. 30. -2xy_, 3i_ 2 « + 10 cM

X- + 2/2 a — 8 6

§ 228; page 202.

12. Vll a62. 14. V2a'3m. 16. ^2 xhn\

13. \/5xi^3, 15. V3m%3. 17. v'3a2x.

22

ALGEBRA.

§ 229 ; pages 202, 203.
19. 6 ab^VS ab-^ + 2 a^. 22. (a; + 3)V5x.

20. 3x?/\/5a;2i/2-4y*.

21. (a-2b)Va + 2b.

27. 12 V6. 29. 42\/2.

28. 5VI05. 30. 75 V3.

35. 12v/50.

23. (3 a - 2 &) \/3a6.

24. (x - 3) Vx^ + 7 X + 10.

31. 28\/42. 33. 7v^.

32. 5\^. 34. 14\/28.

36. 315a&\/l5a&.

2. i\/6. 4.

3. ^\/5. 8.
12. 1-^54.

16. —^420.
Ga

19. ^VSxyz.
9 28

22. -

§ 230 ; pages 203, 204.

iVl5 6. xV^^- 8- iv^

|\/2. 7. 1V34. 9. i^/2.

13. 4\/iO. 14. i\/8T.

1

17. — i-V30x.
10x2

20. i\/98a2.

10. iv^.

11. f\/5.
15. ivTS.

18. -^V22ad.

la

— Va2 _ 62.
-6

23.

x + 2

21

■V2x.

4y

14. VI - a2.

a + 6

17. J(i^

^ X2 + 1

3. 7\/3.

9. 5V3.

14. 0.

18. - a"-h^"

21. ^\/l4.

233; pages 205, 206.

5. -2Vb. 6. 5V^. 7. 3v/3.

4. 4%/2. 5. -2Vb. 6. 5v^. 7. 3v/3. 8. - v^.

10. V7-2\/Il. 11. V-v^- 12. iVe. 13. |v^6.

15. j^VlO. 16. 2v^-3\/5. 17. -4 Vis.

^2ab. 19. 10 ??i2 '^^J^^. 20. (5a - 4x2) V2 a2 - 3x.

22. v^. 23. 6^-2v'6. 24. -3^3.

25. 7v^-5V5.
28. iV3-V6.

26. 4xV0x. 27. 2 62Vl0a& - 3aV76.

30. (7 X - 1) V5x.

31. 13 2/V3.

29. ^|V30-|V10.
32.

a — b

Va^ - 62.

ANSWERS.

23

2. y/27, v/25.

6. \/256, 'C/2I6.

9. %/6i, v^512, v/l69.

§ 234; page 207.

3. ^^2, V9. 5. '^yi28, 5^/144.

8. '-^^810^, v^868, v^36c^.
10. \/l -3x + 3a;--a;3, \/l + 2 a; + x^.

11. Va3 + 3 a26 + 3 afi^ + b^ Va* - 4 a^fe + 6 a262 _ 4 afes + &*,

12. \/3. 13. \/5. 14. v^. 15. V6 > \/l4 > \/l75.
16. v/253>V3>v^. 17. V3 > \/5 > v^.

§ 235 ; pages 208 to 210.

4. 12. 5. 6 a. 6. 6v7. 7. 5V30. 8. 110. 9. 10aV21bc.

10. 12. 11. 3v^. 12. 6^55. 13. |VT5. 14. 3\/l5.

15. 2v^. 16. nxVSx.. 17. 2v'486. 18. v^500 a%x*.

19. 5v^. 20. 2&\/l6a56c3. 21. 3v^. 22. 2v^.

23. 3\/32. 24. §vlt52. 25. J \/l35. 26. v'o^iecS.

27. 2V3. 28. 2\/l()8. 29. v^. S2. 2 + TVS.

33. 12x - 6 + 16 V2x. 34. 202 - 68 VlO. 35. 54 a - 55 6 + dQVab.

36. 1G5 + I8vl0 + 35^100. 37. a - 4 6 + 9c - 6Vac.

38. 22 X + 2 - 23 Vx'- - 1. 39. - 2 - 2\/r5. 40. - 72 + 33\/3.

41. 8 + 30VT5. 42. 140-48V10. 43. -48 + 54 V6 + 12\/l0 + 60\/l5.

44. -47 -2Vl5 + 25\/6. 45. 61 + 24\/5. 46. 37 - 20v'3.

47. 168-96V3. 48. 665 + TOVTO. 49. 5a - 4 + 2V6a2 - 8a.

50. 13x + 5y- 12Vx2 - 2/2. 51. _ 31. 52. 28. 53. 4-21x.

64. 2 b. 55. 3 -46 a.

§ 236; page 211.
3. 2V3. 4. |V5. 5. hVl. 6. 3y/3. 1, -. 8.

9. 3.
14. ^

1 V

10. -^Vl62a3.
3a

11. 2\/2.

12. V2.

56 X
3

15.

16

. /oV^. 17. ^

81

/225,

13.

5c
18. ^^^/4M.

' 343 m5

19. \/96 ax2. 20. iv'leo. 21. a/-- 22. ^^^. 23. vl.

^2 >' 22/

1 .VTTT- «„ 1-/128

24. v/5.

25. -^VlSa. 2_. ^

2o ^243

24 ALGEBRA.

§ 237 ; page 212.

6. 18 v^. 8. a\/Ta. 10. 3v/3. 12. 50m^-^/SVi.

7. S2a'b-Wab. 9. 5V2xy. 11. 2 v^. 14. 2y^'^]^.

§ 238; page 213.

8. v^. 10. v^l62 xy^. 11. </2a. 13. v^. 14. </iTa3.

§ 239; page 213.

2 V6. 4 6v^ 6 )^I^. 8 ^^

■3" '5' ' ' 2 ' ' 2 '

, Vl ab g 4v^9x - v^ « 3\/4x32/5^

7a62 3 3 2x2/3

§ 240; pages 214, 215.

g 9-3a/5 g 53 + 12\/T0 jg 2 a" - ?)2 - 2 g Vg^ - fe"-^

2 " ■ 37 ' ■ fe2
^ 5\/3 + 10 9 22 Vis - 85 j^ 1 - Vl - g^

2 ' 5 'a

g g_+ &^ + 25Vg ^^ .r - 4 - Va: - 2 jg .7: + Vx^ - y\

a — b"^ ' X — 6 y

g x + y - 2\/yy jj b -2g - 2 ■v/g"'^ - g?) jg Vg^ - x* - g^

X — 2/ 6 X-

_ 9 + 4\/3 j2 1 +4xVl -4x2 ^^ 14x-10 + 11 Vx^- 1

3 " ■ 8x2-1 ' ■ 6X-13

§ 241; page 215.

2. .949. 4. .535. 6. - 4.560. 8. .268. 10. -.330

3. 2.224. 5. .684. 7. 4.442. 9. 13.354

§ 247 ; page 218.

3. V7 + 2. 8. 2v^-\/2. 13. 3\/5-l. 18. 2\/5-\/l5.

4. 3-2\/2. 9. 3-\/3. 14. S+VIO. 19. SVS+VlO.

5. 4\/3 + l. 10. \/6+\/5. 15. 5\/2+v'6. 20. 6V2-V3.

6. 2\/3+a/7. 11. 4+v/TO. 16. 3V3-2\/2. 21. Vx+l + Vx-l.

7. 2\/6-2. 12. A/n-3. 17. 4V2-V5. 22. Vg^-Vft.

ANSWERS.

25

§ 251; page 219.

2. 9v^^^. 3. 4V3\/^^. 4. V2\/3

6. {x + y-hz)V^^. 7. 0.

10. V5 V^^. 11. (1 -a;)

1. 5. 6\/^iri.

8. 5V^T. 9. -4a\A^.

§ 252 ; pages 220, 221.

4. -14. 5. 12 a\ 6. -2^15. 7. - Va6. 8. 18. 9. -60.

10. 26-7a/^. 11. G6-33V^^. 12. -Gl + lSVIS.

13. -8a + 18 5. 14. -xyzV^^. 15. 48V2 V^H^.

16. -8\/30- 17Vl5. 17. 2. 18. 480. 19. - V210.

20. -2 + 2\/~^. 21. -74-40^3. 22. U-SV^^.

23. -30 + 12\/6. 24. x2 + ?/ 25. 61. 26. -9a + 4&.

27. - 50. 28. - 12 29. ^ ~ ^~ ^- 30. V^T.

3j 73 + 40V3

23
34. - 10 - 9\/- 3.

29.

32 51 -20VT5
33

33. -2 + 2\/^^.

3. V5.

4. 2.

3. 3.

4. -6,

5.2.

3

6. 16.

» 25
'•36'

8.1.
6

§ 253; page 222.

5. -V^. 7. a/^.

' c

T. 8. ^^

6. 3^

9. ^.

20

10. -2.

11. 2.
12.1

13.

14. 21.
4

254; page 223.

9. \/5.

10. -Vs.

11. \/3„

12. V2.

15. 4.

16. 2i.

17.

18. - 1.

25^
144
_3^
16*

19. --•

3

20. 1.

21.^.
4

22. ^' + ^b\

4 62

23.1^.
3

24. -5.

25. -I«.

26. ^.

27. ?«.
16

28. ^.
64

29. -^.

4

30. -11=

120

31. b.

26 ALGEBRA.

§ 256 ; page 225.

3. ±3. 7. ±5. 11. ±0. 15. ±3^^. 19. ±1.

4. ±|- 8. ±^. 12. ±^. 16. ±1 20. ±ia-b)

5. ±a/3. 9. ±-- 13. ±4. 17. ±8. 21. ± ^VlS.

6. ±6. 10. ±2. 14. ±2. 18. ±i. 22. ± ^•

2 6

§ 259; page 227.

3. 1, -7. 7. 5, -6. 11. -, -5. 15. 6, - -•

2 3

4. 8, - 4. 8. 2, i. 12. 4, -• 16. - -, - -•

3 5 2 3

5. - 2, - 9. 9. -, - 1. 13. -, -•

4 3 3

6. 10, 3. 10. -i, -5. 14. 1, -I.

2' 2 4

§ 260 ; page 229.

3. ^ -2. 7. --, -3. 11. --, -1 15. -, -i-
4 5 5 5 4 5

4. -^, _5. 8. ^, 1. 12. -1, -1. 16. 1 -1.

44 66 62 93

5. 1, ?. 9. 1, -I. 13. -1,-1.

9 8 8 8 4

c 1 3 ,-4 3 1^ ^ 7

6. -J • 10. -, 14. 4,

2 4 7 7 3

§ 262 ; pages 230, 231.

3. 10, -3. 7. -1, -1- 11. --, --• 15. 5, -.

5 2 2 2 5

4. 2, --. 8. 4, -— . 12. -, --. 16. 3,

2 '5 3 5 9

18. 12. 2, _2. ig 3 5^
5 3 5 9

5. y, _3. 9. ,, |. 13. 1. -4. IT. 1, -|.

6. -1, _§. 10. ^, -?. 14. --, -2. 18. -1, -^.

4 2 2 3 6 3 3

ANSWERS. 27

§

263 ; pages 231 to 233.

1.

5, -I

2

10.

•3,

4
5"

20.

-4, -7.

30.

15 2

2.

5 2
6' -^•

11.

1,

7
18

21.

1 1

9' 8*

31.

-1, -2.

3.

3 1
i' 3*

12.

2,

1
3'

22.

-3, -4.

32.

5 ^

4.

-§, -10.
5

13.

3,

1
7'

23.

1 1

2' I

33.

-1,-3

5.

1, -e.

14.

1.

2
3'

24.

4, -1.

34.

.-1

6.

- -3.

15.

2,

-1.

25.

-3, -4.

35.

1 2

5' 7°

7.

2, ?.
' 9

16.

26

2.

26.

'•^•

36.

8 18
5' ~T'

8.

5, -6.

17.

6,

-.3.

27.

f.-3.

37.

'i'-

9.

6±V3.

18.

119, 7.

28.

o 23
^' "12"

38.

h-^-

19.

3,

-13.

29.

o 19,
3, --.

§ 264; pages 234, 235

3. 2a ±3 5. 4. 1, -2m-l. 5. a, -1. 6. -6, -a.

7. w2, m3. 8. ^, -^. 9. 3a + 5, -a + 7. 10. 1, ^^

c a a — 6

11. ^^, _6. 12. -a±V^^^. 13. ^ + 5, a + b

a + b a 2

14. 2 a, - rt. 15. a, - ^?^-+i. 16. a -2b, -86

2a

17 a~b a + 26. ig ^ 13«^ jg 2-^ _2a-l

2 »* 5 2

20. (3a + 6)2, _(3a-6)2. 21. ^^^, ^^. 22. a-26, -2a + 6,

a—b a+b

23. 3a, -li?. 24. a-c, -b + c. 25. 42a, 2a. 26. -— , -2m,

27. 3 a, -4 a. 28. 1, '-:^. 29. ^^tl, ^^i=i

a - 6 a2 - 1 a^ + 1

30. «^^ -^-o 31. a + 6, -2£^. 32. 1, ^ + ^-^2a.

c o — 6 a + 6 c + a — 26

28

ALGEBRA.

§ 265 ; page

236.

3.

3

2'

-2.

7.

1, -I 11.

~h -•

15. -?, 1.

9 6

4.

9,

-4.

8.

4, -1 12.

1 5
4' 4"

1R 1 3
'^- -6' 4'

5.

-

6, -

8. 9.

3, 5. 13.
4

2 4
5' 5"

6.

9

3
5*

10.

-1,-1 14.

14 2

1 7
3' 4'

§ 267 ; pages 237, 238.

5.

-

7, 4.

6. 5,

9. 7. -8, -

3. 8. 12,

-6. 9. 0,

7
5

10.

0,

-8.

11. 0, ±-. 13. 0,
4

5 2
4' §■

15. 3, -^, -

2

-4.

16.

0,

2±-

v/2T.

17. ±3 a, ?, -

- a. 18.

2

^

19.

8,

_3±3V-3
2

n. ±1 ±|v:

n. 21. H

3

-2rt±2a\/-
3

-.3

22.

-

5 5±5\/-3
2' 4

23.

d=2, liV::

~s, -i±V^

^.

24.

1,

±v

- 1.

25. 0, -. 26. ±5, -.

2 5

27. ±1 -

5
'2

28.

-

1-

3V-2.

29. 0,

4a
5

30. 0, 4a-

4.

§ 268; pages 239 to 242.

3. 21, at ^6 per barrel. 4. 11 and 7. 5. 9 and 16 ; or, - — and - — .

2 2

6. 3, 4, 5. 7. 16 and 4 ; or, 25 and - 5. 8. 5 and 2.

9. 1, 2, 3, 4 ; or, 5, 6, 7, 8. 10. 18, at $6 per barrel. 11. 21.

12. $40. 13. 4,5,6. 14. 6 miles an hour. 15. 300.

16. 27 and 36 miles an hour. 17. 18 rods, 12 rods. 18. 20 cents.

19. $ 75 or $ 25. 20. 9 miles an hour.

21. Area of picture, 25 sq. in. ; width of frame, 4 in.

22. Fore-wheel, 12 ft. ; hind-wheel, 16 ft.

23. Larger, 6 hours ; smaller, 10 hours. 24. 22. 25. $3000.
26. 5. 27. 5. 28. 8. 29. 4.

30. 12100 and 1225 sq. ft. ; or, 88.36 and 4489 sq. ft.

31. 6. 32. 136 or 68 miles. 83. 72 miles. 34. 80, at $ 60 each.

ANSWERS.

29

4. ±3, ±2V3.
8. ^,-1

§270; pages 244, 245.

5. ^, ^^. 6. 4, v'm.

10 4

r.-i,-i.

9. - 243, 20'.

') 3 li

11. 729,

76

12. -2,

8\2n

15. V8, (-?)

1 8

13, ±i, ±--

2 4

16. ±— , ±— .
27 125

18. 04,

19. - , f-V.
250 V5,

22.

i' (I)'

23. a,

^2

20. 1,

24. 10, 9.

U. :t?f, ±2-1

17. ±2, ±3.

21. ?56, 2f.
81

25.

49 ±\/97

4. 6, - 3, 3, - 1.

271 ; pages 246, 247.

5. 3, - 7, 1, - 5.

6. 6, - 1, 4, 1.

7. ±3, ±3\/2. 8.0,—- 9. 1,1±2\/15. 10. 0, -2, -1±2V'^.

11. 1 1 3±^^-503. 12.4 1. i3._2,-l^. 14.-2,-5,-3,-4,
'2 i 8

15. 2, - '^, ^^y^. 16. 2, - ^, 1, -■ 17. -, - 2, ?, - 1.

2 4 2 2 3 3

18. 2, l,9±yiM. 19. a±Va'-ib\ gO. a±3 6V36, a±26V2&.

4 8 2

21. 3, -1, 2±^. 22. ^-±^, 9±^^^3. 23. 7, -??, ^±4^2.

2 2 0 5 5

24. 3, -;j^-^.

>'128

25. 5 a, — 7 a, a, — 3 a.

§ 276 ; page 250.

2. X = S, y = ± 5 ; or, X = — 3, ?/ = ± 5.

7 9 7 9

3. X = -, w = ± - ; or, x = , ?/ = ± —

2^ 2' 2-^2

4. X = 2>/8, ?/ =: ± 2 v'2 ; or, x = - 2\/3, y = ± 2\/2,

5. X = 2 a - 6, y = ±(2b + a); or, x = - 2 a + 6, y = ±(2b + a).

30 ALGEBRA.

§ 277 ; pages 250, 251.

Note. — In this, and the three following sections, the answers are
arranged in the order in which they are to be taken ; thus, in Ex. 2,
the value x = 2 is to be taken with y = S, and x = 10 with y = — 13.

7. X = 6, 1.

y = i, e.

8. X = a + 1, ~ a.
y = a, — a — 1.

9. x = 8, -3.

y = 16, -•
^ 3

10. X = a + b, a — b.

y = a — b, a + b.

11. X = 5, - 3.
^ ' 3

§ 278 ; page 253.

4. X = 8, 6. 9. X = 5, 2. 14. x = 8, - 2.
2/ = 6, 8. y=_2, -5. j/=-2, 8.

5. x = l, -10. 10. x = -l, -6. 15. x = 6, -9.
y=-10, 1. y=_6, -1. y = 9, -6.

6. x = 4, -3. 11. x = 5, -7. 16. x = 4, 17.

y = 3, -4. y=_7, 5. 2/=_i7, _4.

7. X = 5, - 9. 12. X = 2, - 16. 17. x = ± 7, ± 13.
y = 9, - 5. 2/ = 16, - 2. J, = ip 13, ^ 7.

8. x=±6, ±2. 13. x = 4, 20. 18. x = 2, -7.
y = ± 2, ± 6. 2/ = _ 20, - 4. y=-7, 2.

19. X = - 6, - 25.
y = 25, 6.

§ 279; page 254.

2. x = ±4, ±|\/2. 3. x=±2, ±f\/2.

y-±l, T|V2. J/=T5, T jV2.

2.

x = 2, 10.

2/ = 3, - 13,

3.

x = 6, -9.

y = - 9, 6.

4.

X = 8, - 7.

y = 7, -8.

5.

X = 10, - 3.

y = 17, 4.

6.

x=2, -5.

2/ = 5, - 2.

12.

. = -4,|.

y=-3,^L3,

9

13.

x = 2,|.

2/-^. 2.

2

14.

. = -.A.

.=3,-?I.

15.

x-4 -^.
X_4, ^

10 12

2/ = 12, -y

ANSWERS.

31

4.
5.
6.
7.

5. X

y

7. X

y

10. X

y

13. a; = 4,

y = o,

16. a;

y

19. X
23. X.

y

26. X

2/
28. X

x = ±S, ±4V3.

y=±6, T5V3.

x = ± 4, ±1.

2/=T2, t3.

x=±Q, ±V-^^3.

2/=T4, ±J/-x^.

x=±4, ±fV7.

y=±S, T\V7.

8. x = ±2, ± j%V- 13.
2/=Tl, if'W-lS.

9. x = ±5, ± |V31o.
y =± 1, T |V- 10.

10. x = ±l, ± fV??.
y=T7, ±fV77.

11. x = ±2, ±x*tV§.,
2/ =±5, :ffiV3.

-±4 3.
-^3 4.

-±4, ±iV^

= ±3, TiA/^

= 5, -4.

= ± 4, ± J a/46.

5,8,1-^
2 2

8, - 5, 16.

= ±2, ±V-T.
= Tl, ±2V^.
= a ± 1. 20. X

= « T 1. y

= 2a-S,

-3a-2,

280 ; pages 257, 258.

6. X = 4, - 3, - 1 ± VT3.

y = 3, -4, 1±\/13.
19

8. X=:±l, ±

y^T3, ±

11. x = 8, 11.
2/=-ll, -

14. x = 2,

2/ = 4,

17. x^3.

15. x = 2, -1,

9. X = 3, 6:

2/ =-6, -3.

12. x = 3, 9.

2/ = 9, 3.

1±\/^^

2/ =-1,2,

1TV^T5

-6. I -I.

18. x=-5, -

55

1, 1, -'-,?•
4 4

10 q
13 '
126 a -169

y-
1 1

3' 4"
.1 1

■4' 3

24. x=±3, ±1

ft 58
y=-6, -y

21. x=i6, -6. 22. x = a±b

y=±3, ±3. y = aTb

25. x=3a + 2,2rt-3

26

±2, ± if V^^31.
T 2, ± liV^TIl.
3, 2, 5zbV^15T

2/=±l, ±3. ?/=2a-3,3ffl + 2

27. x = ±(2rt- 6), ±(a-26)
2/=±(a-2&), ±(2a-6)

29.

„ 10 5 ± V193

X = 2, , —^

3 4

y=-2, -3,

-5±V- 151

2/ =-5, -21,

63±3\/l93

32 ALGEBRA.

30. x=27, -8. 31. x = 1, -1. 32. x=a + l, -a. 33. x=2, 12.

j/ = 8, -27. 2/=-l, 2. 2/ = ffl, _a_i. j,= _3, _i_.

72

34. X = — , - — , 0. 35. X = 4, - 1. 36. X = 2 a, - a.

2 4 5

y=%%^- 2/ = 2, 1. y = 26, -6.

2 4 5

37. x = 2, -1, l±:^ElI. 38. x=0, 2, ± ^. 39. x=±3, ± V^.

2/ = l, -2, -1±^^^^. 2/ = 0, 2, 2tV2. J/ =2, 6.

40. x=±l,±2. 41. x=3, -1, -1, -2. 42. x = 3, 4, -6± v/43.

2/=±|±| 2/ = l, -3, 2, 1. 2/=_4, -3, 6±V43.

43. X = - 2, - 4. 44. X = 3, -1,2, - .3. 45. x = 2, 1.

2/ =-4, -2. ?/=-!, 3, -3, 2. J/ = 1, 2.

46. x = 2, -1^; ,=-1,-29; .=_4,-l-l.
4 6 20

§ 281 ; pages 258 to 260.

1. 6, 4 ; or, -6,-4. 2. ± 5, ± 3 ; or, ± 3\/^ T SV^H^.

3. 18 rods, 9 rods. 4. 7, 5 ; or, - 5, - 7. 5. 5, 2.

6. Cow, $70; sheep, $40. 7. 32 or 23. 8. 9, 4. 9. - or — .

' 8 22

10. 24 m., 16 in.

11. Rate of crew in still water 6 miles an hour, of stream 3 miles an

hour ; or, rate of crew in still water ^-^^ miles an hour, of
stream | miles an hour.

12. Length 30 rods, width 12 rods ; or, length 60 rods, width 6 rods.

13. 60 ; A gives to each !^3. 14. A, 6 hours ; B, 3 hours ; C, 2 hours.

15. Length 32 rods, width 30 rods.

16. 6 and 4 ; - 4 and - 6 : or, ^ ^ ^^^^ and ~" ^ ^^^'^.

2 2

17. A's rate of walking, 3 miles an hour ; distance 12 miles.

18. A, 4 hours; B, 8 hours ; C, 12 hours.

19. 1 and 3 ; or, 2 + Sa/^H^ and 2 ^ SV^H^.

ANSWERS.

33

§ 283 ; pages 262, 263.

2. a;2- 15x + 54 = 0.

3. a;2 + a; - 6 = 0.

4. 3 x2 - X - 2 = 0.

5. 2 x2 + 19 X + 44 = 0.

6. 30x2-31x + 5 = 0.

7. 28 x2 - X - 15 = 0.

8. 8x2+ 17x = 0.

9. 36 x2 + 77 X + 40 = 0.

10. x2+(2 6-3ffl)x+2a2-5a6-3&2=o,

11. x2-2ax + a2-9m2 = 0.

12. x2 - 6 X - 89 = 0.

13. 4 x2 + 4 X Va + a - 6 = 0.

285; pages 264, 265.

9-4x)(5 + 3x).
7-2x)(6 + 5x).
;6x-5)(4x- 1).
;4x + 5)(2x4 7).
3x-4?/)(7x + 6y).
7x- 5a5)(x + 6«6).
x-3y + i)(x + 4y + 3).

2y-l)(x + 2/ + 2).
x-2y + 4)(x + 2?/-l).
2x-2/ + 3)(x + 4y-l).
a-2b -2)(3a + b - 1).
3y-2-x)(3y-3 + 4x).
2 X ~ 5y - z){3x + 3y + 2 z).

§ 286; page 266.

4. (2x + 5)(2x + 3). 5. (3x - 2)(3x - 4). 6. (4x + 7)(4x - 3).

7. (x + l + 2V3)(x + l-2V3). 11. (5x + 3+\/3)(5x+3-V3).

8. (2x + l + v'2)(2x+l-\/2). 12. (2 V2-2 + 3x)(2\/2 + 2-3x).

9. (6x + 5)(6x- 1). 13. (7x + 6)(7x + 2).
10. (x + 2)(4x-3). 14. (1 + 8x)(5-2x).

6.

(3x-2)(x + 3).

19. (

7.

(5x + 8)(x + 2).

20. (

8.

(2x-3)(3x- 1).

21. (

9.

(3x-4)(5x + 2).

22. (

10.

(5-3x)(4 + .x).

23. (

11.

(5-3x)(7 + 2x).

24. (

12.

(6-x)(2 + 5x).

26. (

13.

(x-7a)(3x + 4a).

27. (

14.

(3x -7m)(2x -3m).

28. (

15.

(7x + 2)(2x + 3).

29. (

16.

(3x-2)(6x- 1).

30. (

17.

(l-4x)(5 + x).

31. (

18.

(9x + 2)(2x + 3).

32. (

§ 287; page 267.

5. (x2 + 3x-5)(x2-3x-5).

4. (x2 + 2x + 3)(x2-2x + 3)

6. (2 a2 + 3 a6 + 4 62) (2 a2 _ 3 ab + 4 &2).

7. (3x2 + 4x?/-2?/2)(3x2-4x?/-2?/2).

8. (4 m- + 3 mn + n"^) (4 m^ — 3mn + n^).

9. C2a2 + 5a-7)(2a2-5a-7).

34

ALGEBRA.

10.
11.
12.
13.
14.
15.
16.
17.
18.
19.

:3 x2 + X Vl3 + 3) (3 x^ - a;\/l3 + 3).

2 m^ + mV5 — 2)(2 m'^ — mVu — 2).
>2 + 2 X V2 + 4) (x2 - 2 X V2 + 4) .
;x2 + X V3 - 1) (x2 - xV3 - !)•

3 a- + 5 (T!,r — 5 x'-) (3 rt^ _ 5 ax — 5 .r-).

4 a^ _). Qt)7i + 6 m-) (4 cfi — a??i + 6 ??i-).

5 x2 + X - 2) (5 x2 - X - 2).

5 m^ + 2 mx + 4 x"^) (5 m^- — 2 mx + 4 x'-) .

4 x2 + 2 xy - 7 ?/2) (4 ..^2 _ 2 .,-y — 1y-).

;6 a2 + 2 a& V'2 - 5 &-) ((3 a-^ - 2 a6 V2 - 5 &2).

§ 288 ; page 268.

6. \/3 zfcvTS - V3 ± Vl.5
2 ' 2 '

§ 299; page 277.

3. X = 0, ?/ = 1 ; X = 6, 2/ = 3 ; x = 3, ?/ = 5.

4. X = 4, y = 13 ; X = 8, 2/ = 0. 5. x = 3, ?/ = 5.

6. X = 4, ?/ = 122 ; X = 13, y = 91 ; x = 22, ?/ = 60 ; x = 31, ?/ = 29.

7. X = 3, y = 50 ; X = 10, y = 20 ; X = 17, 2/ =: 2. 8. x = 3, ?/ = 2.
9. X = 3, 2/ = 59 ; X = 13, ?/ = 16.

10. X = 78, 2/ = 4 ; X = 59, 2/ = 12 ; x = 40, 2/ = 20 ; x = 21, 2/ = 28 ;
X = 2, 2/ = 36. 11. X = 2, 2/ = 1, 2 = 3.

12. X = 2, 2/ = 30, z = 3 ; x = 9, ?/ = 18, £! = 48 ; x = 16, ?/ = 6, 2 = 93.

13. x = 2, 2/ = l. 14. x=5, 2/=2. 15. x=8, 2/ = 6. 16. x=3, 2/ = ll.
17. x = 7, 2/ = l. 18. X = 9, 2/ = 4.

19. Either 2 and 8, or 6 and 3, twenty-five and twenty-cent pieces.

20. Eitlier 1 and 17, 3 and 12, 5 and 7, or 7 and 2, fifty and twenty-

cent pieces.

21. Either — and -, — and -, or i and — .
9 5 9 5 9 5

22. Either 1, 18, and 1; 4, 10, and 6; or 7, 2, and 11, half-dollars,
quarter-dollars, and dimes. 23. 5 pi,e;s, 10 sheep, 15 calves.

24. Either 17, 2, and 8 ; or 3, 11, and 25, quarter-dollars, twenty-
cent pieces, and dimes.

ANSWERS. 35

§ 322 ; pages 285, 286.

4. 8. 5. 30. 6. ^. 7. If. 8. ^AJzJ. 9. iqi. iq. ^ - 3.

11. 2a- 1. 12. -1,A. 13. 5,22,-4. 14 ?.

11 b

15. a; = ± a'-^&, 2/ = ± a62. 16. 32,18. 17. 25,11. 18. 31,17.

19. 6, 8. 23. 3 : 4. 24. a : - b. 25. 1 or - 15. 29. 5 : 4.
30. 3:4. 31. 3, 9, 27.

§ 332 ; pages 289, 290.

3. 72. 4. 2/ = 1^3. 5. i. 6. 1^. 7. ^. 8. -18. 9. 1.

3 9 8 4

10. 579 ft. 11. §^, -A. 12 7. 13. 16. 14. H. 15. 12 in.

4 3x 2

le. Z. 17. 5. 18. 9in. 19. 16(A/3-l)m. 20. y = 3 + 5x-ixK

§ 337 ; page 292.

2. 1 = 69, S= 432. 3. Z = - 77, S = - 630. 4. 1 = 36, S = - 264.

5. z=._6£,,^^_561. 6. 1=111,8 = "^.

4 4 4 4

7. ?=!Ǥ, ^^Hli. 8. l = -^,S = -165.

o b 4

33 741

9. 1=-^, S = -^- 10. Z = 34 a + 19 6, 6' = 162 a + 63 6.

6 10

11 ; — 17 .'/ — 8 X ,, _ 80 y — 35 a;

~ 2 ' ~ 2

§ 338 ; pages 294, 295.

4. a = 1,5'= 540. 5. a = 7, ^ = - 09. 6.d = 3,S = 552.

t. cZ=-5, ?=-95. 8. fZ = i,n = .35. 9. a=-,d = -~.

4 5 15

10. Z = — ,w = 16. 11. n = 22,S = -- 12. a = -3, Z = 5.

12 2

13. rt=-^,ri = 9. 14. a=-,d=--- 15. cZ = --,n = 13.
3 2 3 4

16. (Z = -,Z = 6. 17. n = 15, Z = -3; or, n = 6, Z= — .

8 ' ' ' ' 4

18. «=--, ?i = 16; or, a = — , H = 25. 19. n = 16, Z = - 16.

3 ' ' 16

36 ALGEBRA.

21. d = ^^^. 22. ,?=2(;^L:z.a«1^^^2j5-^

71 — 1 n(n — 1) n

23 „^2^- »(?t-l)t; ;^2<S'+n(?i-l)(?

2 ii ' 2 ?i

24 n = ^ ~ ^ + ^ ^^(; + a)(Z-a + d)
' ■ (Z ' 2(Z

25. « = Z - (h - l)d, 6' = " [2 Z - (^i - l)d].

26. a^2^-^^^d = ^("'--^).

n n(n — 1)

27 ^ ^ - f? ± VH d^y + (2 g - dy^ 23 ,; = ^!_=L«!_ „ = ^
2 ■ 2S-a-V a + l'

oa <^± V(2 Z + d)2 - 8 d5 2 ^ + fZ T V(2 Z + d)--^ - 8 dS

<»y. Of = ^ , ?t = ^' ' •

2 2d

§ 339 ; page 296.

2.

-?•

3. d

=

7 A 7 7 -

• 4. a = — 5.

6 18

d:

_3_
4*

6. d = -A

12

7.

.-!•

§ 340 ; page 296.

1.

2 ^
16*

2.

X2

+ 49. 3. 4«' + l.
4 a-'^ - 1

^ ab(a+b)

341 ; pages 297 to 299.

3.

5050. 4. 250500. 5. - 50. 6. 10, 2,

-6, -14.

7.

840. 8. 65 X + 52 ?/. 9. 3, 5, 7, 9. 10. 100.

11. 44650.

12.

31. 13. -. 14. - 6, - 2, 2, 6, 10; or, 21, — , — , -

A 11

12 65

7' 7'

15.
21.

am + h^ 16. 124. 17. 17. 18. 30. 19. 5.

m + 1

-3, 7, 17; or, -3, -i^, -^. 22.579.

5 5

§ 345 ; page 301.

20. 15.

3.

Z = 2187, .9=3280. 4. Z= ^^^, ,9= '*"^. 5./= 1250,

^=-1042.

243

6. Z = 2048, /S = 4094. 7. l=- — ,S^-—-
' 266 256

ANSWERS. 37

8. Z=-1280, ^ = -5Al^. 9. 1=^^, s = -^^-

2 625 625

10. 1 = -^-^, S=-^^. 11. 1 = ^,8 = '-^.

64 192 128 384

12. Z = 768, S = ^^-
4

§ 346 ; pages 302, 303.

3. a = l, aS' = 511. 4. a = 3, Z = — . 5. r = - 4, ,5 =1638.

27

6. n = 10, ^ = '^. 7. a = \ Z = - —

256 2 2048

8. r = ?, ^ = 19111; or, r = --^, ^ = 4039. g ^^1 ^^^

2 384 2 384 2

10. Z = - — , n = 6. 11. a = 3, n = 7. 12. r = -, n = 8.

324 2

13. Z=£±i!JzIM. 14. r = ^^::^. 15. a = W -(r - 1),S'.

r ^ — ^

16. a = -L-,s= ^(r'^-^K n. . = (>-^)'^,z = ^"-V-i)-^.

yn-I j.n-l(-|. _ 1) r" — 1 r" — 1

1 _n_ _n_

§ 347 ; pag-e 304.

2. ?. 4. -'-. 6. 1^. 8. -A.
2 6 5 40

3. ^. 5. -5. 7. 12. 9. A.

5 55 21

§ 348; page 305.

2 A. 3 A. 4 2^. 5 ^. 6 — • 7 ^^^

ll" ■ 27* ■ 36" ■ 990* ■ 925* ' 2475*

§ 349; page 305.

2. J- -8. 3. r=-2. 4. r=±2. 5. ?-=±^. 6. r=-4. 7. r=±--

2 3

§ 350 ; page 306.

1. 21 2. 1. 3. a2-&2. 4. E±ll.

^ x-'2y

38 ALGEBRA.

§351; pages 306, 307.

2. -4. 3. 4, 12, 36, 108. 4. 5, -10, 20; or, -5, -10, -20.

5. $4118. 6. 32 ft. 7. - ^. 8. (a'"b)^^\

256

9. -3,4, 11; or, 13,4, -5. 10. A, $108; B, $144; C, $192; D, $256.

11. - 4, 1, 6, 36 ; or, 8, 1, - 6, 36. 12. 3.

lo ^ « n 76 190 475
13. 4, 6, 9; or, — , — ,

' ' ' ' 39 39 39

§ 355 ; page 309.

3. -A. 4. A. 5. A. 6. -A. 7. -I

13 229 61 17 6

8. 2,10,10, -10, -'A -2, -15, _10.

'3 3 7 9

9. -l-i-l, -4, 2, 4,1, ±,2.

5 7 5 2 11 7

10.

4_3 _12 _2 _12 _3_ _± _6_ _12 _1_
5' 6' 25' 6' 35' lO' 15' 25' 55' d

11. 4. 12. ?— ^- 13. -^^, ^y , -^ 14. 5 and -3.

X 2x—y 3x—2y 4x—3y

15. -I
9

§ 360 ; page 314.

10. aW + 5 a^^^c + 10 a^^c"- + 10 a*b^c^ + 5 rt26i2c4 + biSgS.

11. a;12m ^ 6 ^WmySn ^, 15 3;8m?/6n _|. 20 X^"'lf" +15 a;*'"?/12n + 6 X^^J/l^n + j/lSn.

12. 16a*-32a3 + 24a2-8a + l.

13. ic5+ 10x* + 40x3 + 80x2 + 80x4-32.

14. rt* - 12 a35 + 54 ^252 _ 108 ah^ + 81 6*.

15. 1 + 12 m2 + 60 m* + 160 ?n6 + 240 m^ + 192 m^ + 64 mi2.

17. x* + 5 .x"^" + 10 x^ + 10 x~^ + 5 x~"*' + x-5.

18. cfi - 14 a3 + 84 J - 280 a^ + 560 J - 672 a + 448 a^ - 128.

19. 243 + 405 x3 + 270 x^ + 90 x^ + 15 xi2 + xi5.

20. m'^ + 6 m"i ^ + 15 ?h~3 + 20 m* + 15 ?n"«' + 6 m^^ + m^.

21. 256 a6 - 256 Jx^ + 96 a^x^ - 16 a^x + xt

22. X-IO - f X-8 y* + -V- X-6i/8 - 10 X-*l/12 + J^ X-2^16 _ _1_ j^20.

23. wi2 + 20 wi9x-3 + 150 m6x-6 + 500 m'x-^ + 625 x-12.

ANSWERS. 39

24. 16 aJ + lGa^ + 6 a~^ + a~" + ^\ a-^.

14 12-1. -l 8 _3 6 4 _5

25. X ^ - 7 X 5 y *^ + 21 x^!/ 2 - 35 x^y ? + 35 x^yi - 21 x^y *

26. 16 a"3' - 32 a'^b^ + 24 a'h - 8 a"^6^ + b'^.

__15 3 _9 fi _3 9 _3 1 2

27. 3^2 X ^ — 1^5 x-^m^ + I X •^m^ — f x ^m^ + | a; ■*m~5" — m^.

28. a6 + 16 oT'^ + 96 a'^'" + 256 a^ + 256 ai

29. a3 _ 18 aV^ + 135 a^x'^ - 540 a^-* +1215 ax"'^"- _ 1453 a^x"'^^

+ 729 x-8.

30. 32 a5 _ 240 a*b + 720 a'^b"- - 1080 a-ft^ + 810 ab* - 243 ft^.

31. a'^b'i + 7 a^6"3 + 2I a^fc-i + 35a^fc~3^ + Sba'h^ + 21 a"^5

-5 5 _7 7

+ 7a -63 + a 263.

32. 81 m^n-^ - 216 mir'^ + 216 - 96 m-hi + 16 m-2n2.

34. 1 - 4 X + 10 x2 - 16 x3 + 19 x* - 16 x^ + 10 x^ - 4 x" + x*.

35. x8 + 4 xT + 14 x6 + 28 x5 + 49 X* + 56 x^ + 56 x2 + 32 X + 16.

36. 1+ 12x + 50x2 + 72x3 -21x*- 72x5 + 50x6 -12x^ + x8.

37. x8-8x^+ 12x6 + 40x5-74x^-120x3 + 108x2 + 216x + 81.

38. l + 5x+5x2-10x3-15x* + llx5 + 15x6-10x^-5x8 + 5x9-xi'>.

39. xio - 5x9 + 20x8 - 50 x'^ + 105x6 - 161 x^ + 210x* - 200x3

+ 160x2 -80x + 32.

§ 362 ; page 316.

2. 56a5x3. 7. -\%^-a-^b\ 12. j%\%aH-^

3. 165 m3. 8. -220xi5y-3. 13. 42240 a'^x^^'.

4. 126a56*. 9. 5005a6m+9n. 14. 21840 m'^^n"^.

5. -11440x9. 10. -219648x-62/i 15. - ^i^fx'^^'y'^.

6. 495 7n8n24. n. 61236 a""3^'x25. 16. -iffa^x^i.

§ 371 ; page 323.

3. 1 + 4x-4x2 + 4.-c3-4x* + •••.

4. 3 + 10 x + 40x2 + 160x3 + 640x4 + ....

5. 2 + 13x2 + 39x4 + 117x6 + 351x8+....

6. 2 X - I x3 + -2^ x5 - -6/ x^ + W ^^ •

7. 1 + x + x3 + 2x* + 5x6h .

40 ALGEBRA.

8. 2x -7x2 + 38x3 -204x* + 1096x5 .

9 1 /V.-2 I 5 /y-l _L 2 5 I 12 5 -V. I 6 2 5 7.2 I ...

in 1 It 11 ^2 25 -J.3 113 T.4 _!_..,

lU. 2 — J * — 8" •*" T6 ■*' I 3 2 •*' T^ •

11. 1 -2x + x^ + 2x3-3x*+ -.

12. 2 + 9x + 23x2 + 47x3 + 73x1+ —.

13. x-3 + 5x-2 + 20x-i + 106 + 570x + —.

14. 3x-2 + 14x-i + 39 + lOlx + 264x2 + ....

15. 1x2-2x3 + fx* - |x5 + ix«+ .".

1(t 214/v. 19 -1.2 5 6-V.3 148/y.4_...

AD. f-t-gX jyJ' sT'*' 2?3-*' •

17. |X-1 - f X + |:X2 + 3 a;3 _ 5 a;4 + ....

§ 372 ; page 324.

2. 1 +2x-2x2 + 4x3- 10x* + ••..

31 St 2 5/V.2 125 /r3 3125 /7.4 ...
• J^ ■jX g-X 16"'*' 128'"'' — •

4. 1 + X-X2 + X3- fX*+ •••.

.*» 1 _ 1 ->• _ 5 /y2 5 T.3 _ 4 5 /V.4 ...

O. 1 -jX jX T6''' T2 8'*' •

6. 1 + X - X2 + f X3 - -LO X* + "..

7. 1 — gX + fX-'+gyX + 23"3''' "I" ***•

§ 374; page 325.
3.^^^ + -^. 4. A 2 5 3_. 1

2x + 3 2x — 3 3x 3(5x — 6) x x+5 x — 6

6 8 I 7 Y 4a 3a g 10 ^ 3

2x + 3 3x — 2 x + 5a x — a 2 — 5x 4 + x

9. 1 +— ^ 1_. 10. 1-^- + . 2 1

2(2x-l) 2(4x-3) 3x + 2 x x-2 x + 3 x-3

376; page 327.

2 13

2. 6.

2x-3 (2x-3)2 5(5x+2) (5x+2)2 5(5x + 2)S

3.-1 «— +-: 4_. 7. 1 4 ^

X + 5 (x + 5)2 (x - 5^3 x - 1 (X - 1)3 (X - 1)*

4._i § _^- - 8. -J 2_^ 3 4

3x-l (3x-l)2 (3i^-l) x+2 (x + 2)2 (x + 2)3 (x + 2)4

5 _^+_i^ I . 2 +- 5 4

2x-3 C2a;-3^2 (2a;-3)s 3(3x-2) 3(3x-2)2 3(3x-2)4

ANSWERS. 41

§ 377 ; page 328.

2. 2 5 8^ 5 1 . i _ _3 ^_

■ X X - 3 (x - 3)2* ' X x^ X - 1 (x - 1)2

x x2 x3 X + 4 ' X X + 1 (x + 2)2*

4. ^ + -^ ^ 7.-^ ^ I 5

3x-l 2x + 3 (2x + 3)2 4x+l 2(2x-3) 2(2x-3)2

§ 378 ; page 329.

2.3X-2+-? 2_. 4.x-l-l-2. + l + _i_.

X + 2 3X-1 xx2x3x + l

3.2 ^- + _J 5.X + 2+?-! i- + — 2

x-2 (x-2)3 ^ ^x x2 a;-l^(x-l)2

3 1,2 6

6. x2 + 3 - § - A

+ --

X x2 x3 X + 3

§ 379 ; page 330.

2. -1-+ 3x-l . 5 ^ L_ + ^ril.

X+lx2-X + l X+lX-lx2+l

3. 5 ^ 2x + 3 g ^^ 3x + 1

3x+l X2-X + 3 ■ 2x-3 4x2 + 6x + 9'

4. ^ _ ^-3 Y 5x + 6 _ 3x-4

■ 2x-5 x2 + 2 * x2 + x + 1 x2-x + l'

§ 380 ; page 331.

. 2. X = jr + y2 + y3 + y4+ ..., 6. X=y + |2/2+iy8+ ly4+ ....

3. x = y + \y'i + \y^ + ^\y*+ .... 7. x = 2y - 2?/2 + | y8 _ |^+ ,.,

4. x = y-2?/24.5y3_ i4y4_|. ,,._ 8. x = y-y^ + y^ -y^ + ....

5. X=y + 3j/2^i3y3 + 67j^+ .... 9. x = y-\y^-\-^^y^—^^^^yT+...

§ 383 ; page 336.

7. a^ -\ oT^x - j\ a"^x2 - ^|^ a-V^s _ ^|^ a""'?' ic*

8. 1 — 1 x J- -3^r2 JJ -yS J 44 T.4

9. a-6 -f 6 a-Tft + 21 a-^b"^ + 56 a-^b^ + 126 a-i06*+ ....

10. x^ - 5 xy + -V- a;V - f a;"3y3 _ 5 X-V4. ....

11. m8 -- 1 wiion"^ + V-»»^2"~-' - f^mi*n"^ + ffTn^n"^ - ....

42 ALGEBRA.

12. rt-i + -^ a-''x~^^ + I a-9x-i + if a-i^a;"^ + iff a-i^x-2+ ....

13. x-2 - 4x-*2/ + 16x-6j/2 _ 64x V + 256x-iV •

2 1 15 9 3 3

14. x~ 2" + 7 x~ 2'"?/s + SjS x~"y'Z^ + V xT^y^z^ + s^a a;^2/*2;*+ ....

15. wr^ + 10m-%~3^ + 60m"itt"^ + 280ift-*w-2 + 1120 nr^-n~^ +

33 1111 _1919 _2 727 _3 535

17. X + Sx'?/* + 20x^ y^ + s.^^'^x^y^ + ^^x'^Y+ -■.

_9 _3 2 3 4 9 6. 1_5 8

18. 8 a » - 3 a ^x^ + /^ a%^ + ^ij a«x5 + 3:5^^ a »'x^+ .•••

§

384; pagre 337.

2.

z|3« ^X*.

7.

V#x*.

12.

V#« '^'^"^°-

3.

j%%\a-'h^.

8.

40
6561 " •'' •

13.

_ 14^8_0 x'^3y-3^

4.

1366x11.

9.

- 2002 x-i5m9.

14.

220 x-i32/""'2''2-6.

5.

- 192 x''y\

10.

64|5 TO-Vn-28.

15.

- ^IF a"'^'6-*.

^^* J176 8 * " ^ •

§ 385 ; page 338.

2. 5.09902. 4. 2.08008. 6. 2.03055.

3. 9.89949. 5. 2.97182. 7. 1.96100.

§ 397; page 342.

2. 1.5441. 7. 2.1003. 12. 2.5104. 17. 3.0512.

3. 1.6990. 8. 2.2922. 13. 2.5774. 18. 3.4192.

4. 1.6232. 9. 2.3892. 14. 2.6074. 19. 3.7814.

5. 1.8751. 10. 2.3222. 15. 2.9421. 20. 4.0794.

6. 1.6020. 11. 2.7960. 16. 2.8363. 21. 4.2006.

§ 399; page 343.

2. .5229. 5. 1.6532. 8. .2831. 11. 1.4692.

3. .2431. 6. .2689. 9. .7939. 12. 1.3468.

4. 1.1549. 7. 2.3522. 10. 2.1303. 13. 2.0424.

§ 402 ; page 344.

8. 3.3397. 5. .7525. 7. 7.7205. 9. .2863.

4. 4.1940. 6. .6338. 8. .4824. 10. 1.0460.

ANSWERS.

48

11. .3943.

12. .0682.

13. .1165.

14. .0939.

15. .4042.

16. .6250.

17. .4978.

18. .2542.

20. .0495.

21. .0365.

22. .7007.

23. .8752.

24. .0794.

25. .4248.

26. .1341.

27. .1807.

§ 406; page 346.

2. 0.4471 6. 1.5104. 10. 6.5353-10. 14. 3.2646.

3. 1.0491. 7. 7.5741-10. 11. 9.9421-10. 15. 0.1151.

4. 9.7993 - 10. 8. 3.8293. 12. 0.4134. 16. 0.7336.

5. 8.9912 - 10. 9. 8.5932 - 10. 13. 2.4383.

§ 411 ; page 350.

6. 3.0286. 9. 7.8605 - 10. 12. 2.4032. 15. 7.8108 - 10.

7. 1.9189. 10. 0.8923. 13. 9.9632-10. 16. 8.1332-10.

8. 9.9830-10. 11. 6.5783-10. 14. 3.6099. 17. 0.6059.

4. 64.26.

5. 2273.

6. 461.2.

1. 189.7.

2. 8.243.

3. -1933.

4. .3091.

5. .002976.

1 6. -.01213.

,7. 6.359.

, 8. .03018.

9. -5.853.

10. 311.9.

11. .2239.

12. -.009544.

13. .1261.

14. .02367.

§ 413 ; page 351.

7. .8143. 10. .09215.

8. .004897. 11. 64.23.

9. 7.488. 12. .003856.

16. .0001994.

§ 418; pages

15. -1.167.

16. -.002893.

17. 3692.

18. .2777.

19. -15893.

20. .001233.

21. 316.2.

22. .7652.

23. 243.9.

24. .00001085

25. 2.236.

26. 1.149.

27. - 1.276.

28. 1.778.

355, 356.

29. .6682.

30. .6458.

31. .1377.

32. -.3702.

35. 30.12.

36. 2.487.

37. 1.056.

38. .0006777.

39. .007105.

40. .8335.

41. .5428.

42. - 36.03.

43. -11.11.

44. .9432.

13. .5061.

14. 356.8.

15. 17008.

45. 2.627.

46. 2.527.

47. -.8378.

48. 1.033.

49. .2984.

50. .3697.

51. .7945.

52. .9348.

53. 179.5.

54. 1.883.

55. .0001931.

56. -.09954.

57. .1711.

58. -74.88.

44 ALGEBRA.

§ 419 ; page 357.
3. .28301. 4. -2.172. 5. 1.155. 6. -.1766.

7. ''^SC 3 31oga g 1. ^^ _^

log a — 2 log b log n — 4 log m 2

11. ^^log^-logg I ^ 12. ^-log[(r-l)/y+a1-loga

log r log r

13. n = \ogl-'ioga + i_

log(^-a)-log(/S-0

14. ^_log?-IogrW-(r-l)^1 J i_

logr

§ 420 ; page 358.
2. 3.701. 3. -.06552. 4. -2.761. 5. 2.389.

6. -.3763. 7. .3731. 9. 4. 10. ^. 11. -h 12. 1

3 3 5

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